Probability and Statistics
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AVU-PARTNER INSTITUTION
MODULE DEVELOPMENT TEMPLATE
PROBABILITY AND STATISTICS
Draft
By
Paul Chege
Version 19.0, 23rd March, 2007
C.
TEMPLATE STRUCTURE
I.
INTRODUCTION
1. TITLE OF MODULE
Probability and Statistics
2. PREREQUISITE COURSES OR KNOWLEDGE
Secondary school statistics and probability.
3. TIME
The total time for this module is 120 study hours.
4. MATERIAL
Students should have access to the core readings specified later. Also, they will need a
computer to gain full access to the core readings. Additionally, students should be able
to install the computer software wxMaxima and use it to practice algebraic concepts.
5. MODULE RATIONALE
Probability and Statistics, besides being a key area in the secondary schools’
teaching syllabuses, it forms an important background to advanced mathematics at
tertiary level. Statistics is a fundamental area of Mathematics that is applied across
many academic subjects and is useful in analysis in industrial production. The study
of statistics produces statisticians that analyse raw data collected from the field to
provide useful insights about a population. The statisticians provide governments
and organizations with concrete backgrounds of a situation that helps managers in
decision making. For example, rate of spread of diseases, rumours, bush fires,
rainfall patterns, and population changes.
On the other hand, the study of probability helps decision making in government
agents and organizations based on the theory of chance. For example:- predicting
the male and female children born within a given period and projecting the amount of
rainfall that regions expect to receive based on some historical data on rainfall
patterns. Probability has also been extensively used in the determination of high,
middle and low quality products in industrial production e.g the number of good and
defective parts expected in an industrial manufacturing process.
2
II.
CONTENT
6. Overview
This module consists of three units:
Unit 1: Descriptive Statistics and Probability Distributions
Descriptive statistics in unit one is developed either as an extension of secondary
mathematics or as an introduction to first time learners of statistics. It introduces the
measures of dispersion in statistics. The unit also introduces the concept of
probability and the theoretical treatment of probability.
Unit 2: Random variables and Test Distributions
This unit requires Unit 1 as a prerequisite. It develops from the moment and moment
generating functions, Markov and Chebychev inequalities, special univariate
distributions, bivariate probability distributions and analyses conditional probabilities.
The unit gives insights into the analysis of correlation coefficients and distribution
functions of random variables such as the Chi-square, t and F.
Unit 3: Probability Theory
This unit builds up from unit 2. It analyses probability using indicator functions. It
introduces Bonferoni inequality random vectors,, generating functions, characteristic
functions and statistical independence random samples. It develops further the
concepts of functions of several random variables and independence of X and S2 in
normal samples order statistics. The unit summarises with the treatment of
convergence and limit theorems.
Outline: Syllabus
Unit 1 ( 40 hours): Descriptive Statistics and Probability Distributions
Level 1. Priority A. No prerequisite.
Frequency distributions relative and cumulative distributions, various frequency
curves, mean, Mode Median. Quartiles and Percentiles, Standard deviation,
symmetrical and skewed distributions. Probability; sample space and events;
definition of probability, properties of probability; random variables; probability
distributions, expected values of random variables; particular distributions; Bernoulli,
binomial, Poisson, geometric, hypergeometric, uniform, exponential and normal.
Bivariate frequency distributions. Joint probability tables and marginal probabilities.
3
Unit 2 ( 40 hours): Random Variables and Test Distributions
Level 2. Priority B. Statistics 1 is prerequisite.
Moment and moment generating function. Markov and Chebychev inequalities,
special Univariate distributions. Bivariate probability distribution; Joint Marginal and
conditional distributions; Independence; Bivariate expectation Regression and
Correlation; Calculation of regression and correlation coefficient for bivariate data.
Distribution function of random variables, Bivariate normal distribution. Derived
distributions such as Chi-Square. t. and F.
Unit 3 ( 40 hours): Probability Theory
Level 3. Priority C. Statistics 2 is prerequisite.
Probability: Use of indicator functions. Bonferoni inequality Random vectors.
Generating functions. Characteristics functions. Statistical independence Random
samples. Multinomial distribution. Functions of several random variables.
The independence of X and S2 in normal samples Order statistics Multivariate
normal distribution. Convergence and limit theorems. Practical exercises.
4
Graphic Organiser
Variance
&
Standard
deviation
Mean,
Mode, and
Median
Frequency
Curves,
Quartiles
Deciles and
Percentiles,
Indicator
functions
DATA
Bonferoni
Inequalities,
random
vectors
Generating
functions,
characteristic
functions &
random samples
Multinomial
distributions,
Functions of
random variables
Probability
Probability
distributions
Multivariate
distribution,
Convergence &
limit theorems
Moment
and
moment
generating
function
Markov and
Chebychev
inequalities
Joint marginal
& conditional
distributions
Univariate
and
Bivariate
distributions
Regression
& correlation
Derived
distributionsChi-square,
t and F
Joint
probability
tables
5
7. General Objective(s)
By the end of this module, the trainee should be able to compute the various measures
of dispersions in statistics and work out probabilities based on laws of probability and
carry out tests on data using the theories of probability
8. Specific Learning Objectives (Instructional Objectives)
Unit 1: Descriptive Statistics and Probability Distributions ( 40 Hours)
By the end of unit 1, the trainee should be able to:
Draw various frequency curves
Work out the mean, mode, median, quartiles, percentiles and standard deviations
of discrete and grouped data
Define and state the properties of probability
Illustrate random variables, probability distributions, and expected values of
random variables.
Illustrate Bernoulli, Binomial, Poisson, Geometric, Hypergeometric, Uniform,
Exponential and Normal distributions
Investigate Bivariate frequency distributions
Construct joint probability tables and marginal probabilities.
Unit 2: Random Variables and Test Distributions ( 40 Hours)
By the end of unit 2, the trainee should be able to:
Illustrate moment and moment generating functions
Analyse Markov and Chebychev inequalities
Examine special Univariate distributions, bivariate probability distributions, Joint
marginal and conditional distributions.
Show Independence, Bivariate expectation, regression and correlation
Calculate regression and correlation coefficient for bivariate data
Show distribution function of random variables.
Examine Bivariate normal distribution
Illustrate derived distributions such as Chi-Square, t, and F.
Unit 3: Probability Theory ( 40 Hours)
By the end of unit 3, the trainee should be able to:
Use indicator functions in probability
Show Bonferoni inequality random vectors
Illustrate generating and characteristic functions
Examine statistical independence random samples and multinomial distribution
Evaluate functions of several random variables
Illustrate the independence of X and S2 in normal samples order statistics
Show multivariate normal distribution
Illustrate convergence and limit theorems.
Work out practical exercises.
6
III.
TEACHING AND LEARNING ACTIVITIES
9. PRE-ASSESSMENT: Basic mathematics is a pre-requisite for Probability and
Statistics.
QUESTIONS
1) When a die is rolled, the probability of getting a number greater than 4 is
1
A.
6
1
B.
3
1
C.
2
D.
1
2) A single card is drawn at random from a standard deck of cards. Find the probability
that is a queen.
1
A.
13
1
B.
52
4
C.
13
1
D.
2
3) Out of 100 numbers, 20 were 4’s, 40 were 5’s, 30 were 6’s and the remainder were
7’s. Find the arithmetic mean of the numbers.
A.
0.22
B.
0.53
C.
2.20
D.
5.30
7
4) Calculate the mean of the following data.
Height (cm) Class mark (x)
60 - 62
61
63 - 65
64
66 - 68
67
69 - 71
70
72 - 74
73
A.
B.
C.
D.
57.40
62.00
67.45
72.25
5) Find the mode of the following data: 5, 3, 6, 5, 4, 5, 2, 8, 6, 5, 4, 8, 3, 4, 5, 4, 8, 2, 5,
and 4.
A.
B.
C.
D.
4
5
6
8
6) The range of the values a probability can assume is
A.
B.
C.
D.
From 0 to 1
From -1 to +1
From 1 to 100
1
From 0 to
2
7) Find the median of the following data: 8, 7, 11, 5, 6, 4, 3, 12, 10, 8, 2, 5, 1, 6, 4.
A.
B.
C.
D.
12
5
8
6
8) Find the range of the set of numbers: 7, 4, 10, 9, 15, 12, 7, 9.
A.
B.
C.
D.
9
11
7
8.88
8
9) When two coins are tossed, the sample space is
A.
B.
C.
D.
H, T and HT
HH, HT, TH, TT
HH, HT, TT
H, T
10) If a letter is selected at random from the word “Mississippi”, find the probability that
it is an “i”
A.
B.
C.
D.
1
8
1
2
3
11
4
11
ANSWER KEY
1.
6.
B
A
2.
7.
A
D
3.
8.
D
B
4.
9.
C
B
5.
10.
B
D
PEDAGOGICAL COMMENT FOR LEARNERS
This pre-assessment is meant to give the learners an insight into what they can
remember regarding Probability and Statistics. A score of less than 50% in the
pre-assessment indicates the learner needs to revise Probability and Statistics
covered in secondary mathematics. The pre-assessment covers basic concepts
that trainees need to be familiar with before progressing with this module. Please
revise Probability and Statistics covered in secondary mathematics to master the
basics if you have problems with this pre-assessment.
9
KEY CONCEPTS ( GLOSSARY)
1) Mutually Exclusive: Two events are mutually exclusive if they cannot occur at
the same time.
2) Variance of a set of data is defined as the square of the standard deviation i.e
variance = s2.
3) A trial: This refers to an activity of carrying out an experiment like picking a card
from a deck of cards or rolling a die or dices
4) Sample space: This refers to all possible outcomes of a probability experiment.
e.g. in tossing a coin, the outcomes are either Head(H) or tail(T)
5) A random variable: is a function that assigns a real number to every possible
result of a random experiment.
6) Random sample is one chosen by a method involving an unpredictable
component.
7) Bernoulli distribution: is a discrete probability distribution, which takes value 1
with success probability p and value 0 with failure probability q = 1 − p.
8) Binomial distribution is the discrete probability distribution of the number of
successes in a sequence of n independent yes/no experiments, each of which
yields success with probability .p
9) Hypergeometric distribution: is a discrete probability distribution that describes
the number of successes in a sequence of n draws from a finite population
without replacement.
10) Poisson distribution: is a discrete probability distribution that expresses the
probability of a number of events occurring in a fixed period of time if these
events occur with a known average rate, and are independent of the time since
the last event
11) Correlation: is a measure of association between two variables.
12) Regression: is a measure used to examine the relationship between one
dependent and one independent variable.
13) Chi-square test is any statistical hypothesis test in which the test statistic has a
chi-square distribution when the null hypothesis is true, or any in which the
probability distribution of the test statistic (assuming the null hypothesis is true)
can be made to approximate a chi-square distribution as closely as desired by
making the sample size large enough.
14) Multivariate normal distribution is a specific probability distribution, which can
be thought of as a generalization to higher dimensions of the one-dimensional
normal distribution.
15) t -test is any statistical hypothesis test for two groups in which the test statistic
has a Student's t distribution if the null hypothesis is true
10
STATISTICAL TERMS:
1) Raw data: Data that has not been organised numerically
2) Arrays: An arrangement of raw data numerical data in ascending order of
magnitude.
3) Range: the difference between the largest and the smallest numbers in a data
4) Class intervals: In a range of grouped data e.g 21-30, 31-40 etc, then 21-30 l is
called the class interval.
5) Class limits: In a class interval of 21-30, then 21 and 30 are called class limits.
6) Lower class limits (l.c.l) : In the class interval 21-30, the lower class limit is 21.
7) Upper class limit (u.c.l): in the class interval 21-30, the upper class limit is 30
8) Lower and upper class boundaries: In the class interval 21-30, the lower class
boundary is 20.5 and the upper class boundary is 30.5. These boundaries
assume that theoretically measurements for a class interval 21-30 includes all
the numbers from 20.5 to 30.5.
9) Class Interval: In a class 21-30, then the class interval is the difference between
the upper class limit and the lower class limit i.e. 30.5-20.5 = 10. The class
interval is also known as class width or class size.
10) Class Mark or Mid-point: In a class interval 21-30, the class mark is the
21 30
25.5
average of 21 and 30 i.e
2
11) Frequency Distributions:
large masses of raw data maybe arranged in
classes in tabular form with their corresponding frequencies. e.g.
Mass (kg)
10-19 20-29 30-39 40-49
Number of pupils (f) 5
7
10
6
This tabular arrangement is called a frequency distribution or frequency
table.
12).Cumulative Frequency: For the following frequency distribution, the
cumulative frequencies are calculated as additions of individual frequencies
Mass ( X)
20-24 25-29
30-34
35-39
40-44
Frequency (f)
4
10
16
8
2
Cumulative
4
4+10=14 14=16=30 30+8=38 38+2=40
Frequency( C.F)
Hence the cumulative frequency of a value is its frequency plus
frequencies of all smaller values.
The above table is called a Cumulative Frequency table.
11
13). Relative – Frequency Distributions: In a frequency distribution
Mass ( X)
20-24 25-29 30-34 35-39 40-44
Frequency (f) 4
f
10
16
8
2
40
The relative frequency of a class 25-29 is the frequency of the class
divided by the total frequency of all classes (cumulative frequency) and
generally expressed as a percentage.
Example:
The relative frequency of the class 25-29 =
f
f
100% =
10
100 25%
40
Note: the sum of relative frequencies is 100% or 1.
14).Cumulative Frequency Curves ( Ogive)
Mass ( X)
20-24 25-29
30-34
35-39
40-44
Frequency (f)
4
10
16
8
2
Cumulative
4
4+10=14 14=16=30 30+8=38 38+2=40
Frequency( C.F)
12
From the above Cumulative Frequency table, we can draw a graph of
Cumulative frequency verses the upper class boundaries.
Upper class
24.5
29.5
34.5
39.5
44.5
3
14
30
38
40
boundaries
Cumulative
frequencies
Cumulative frequency
Ogive
45
40
35
30
25
20
15
10
5
0
20
25
30
35
40
45
Upper class limit
Note: From the cumulative frequency data, the first plotting point is ( 24.5, 3). If we
started our graph at this point, it would remain hanging on the y-axis. We create
another point (19.5, 0) as a starting point. 19.5 is the projected upper class
boundary of the preceding class.
13
SHAPES OF FREQUENCY CURVES
Symmetrical or bell-shaped.
Skewed to the right ( positive skewness)
Has equal frequency to the left and right
of the central maximum e.g. normal curve
Skewed to the left ( Negative skewness)
Has the maximum towards the right of
the and the longer tail to the left
Has the maximum towards the left and
the longer tail to the right
J –Shaped
Has the maximum occurring at the right
end
14
Reverse J-Shaped
Has the maximum occurring at the left
end
Bimodal
Has two maxima
U- shaped
Has maxima at both ends
Multimodal
Has more than two maxima.
15
COMPILED LIST OF COMPULSORY READINGS
Reading # 1: Wolfram MathWorld (visited 06.05.07)
Complete reference : http://mathworld.wolfram.com/Probabilty
Abstract : This reference gives the much needed reading material in probability and
statistics. The reference has a number of illustrations that empower the learner through
different approach methodology.
Wolfram MathWorld is a specialised on-line
mathematical encyclopaedia.
Rationale: It provides the most detailed references to any mathematical topic. Students
should start by using the search facility for the module title. At any point students should
search for key words that they need to understand. The entry should be studied
carefully and thoroughly.
Reading # 2: Wikipedia (visited 06.05.07)
Complete reference : http://en.wikipedia.org/wiki/statistics
Abstract : Wikipedia is an on-line encyclopaedia. It is written by its own readers. It is
extremely up-to-date as entries are continually revised. Also, it has proved to be
extremely accurate. The mathematics entries are very detailed.
Rationale: It gives definitions, explanations, and examples that learners cannot access
in other resources. The fact that wikipedia is frequently updated gives the learner the
latest approaches, abstract arguments, illustrations and refers to other sources to
enable the learner acquire other proposed approaches in Probability and Statistics.
Reading # 3: MacTutor History of Mathematics (visited 03.05.07)
Complete reference : http://www-history.mcs.standrews.ac.uk/Indexes
Abstract : The MacTutor Archive is the most comprehensive history of mathematics on
the internet. The resources are rganised by historical characters and by historical
themes.
Rationale: Students should search the MacTutor archive for key words in the topics
they are studying (or by the module title itself). It is important to get an overview of
where the mathematics being studied fits in to the history of mathematics. When the
student completes the course and is teaching high school mathematics, the characters
in the history of mathematics will bring the subject to life for their students. Particularly,
the role of women in the history of mathematics should be studied to help students
understand the difficulties women have faced while still making an important
contribution.. Equally, the role of the African continent should be studied to share with
students in schools: notably the earliest number counting devices (e.g. the Ishango
bone) and the role of Egyptian mathematics should be studied.
16
Compulsory Resources
Resource #1
Maxima.
Complete reference : Copy of Maxima on a disc is accompanying this course
Abstract : The distance learners are occasionally confronted by difficult mathematics
without resources to handle them. The absence of face to face daily lessons with
teachers means that learners can become totally handicapped if not well equipped with
resources to solve their mathematical problems. This handicap is solved by use of
accompanying resource: Maxima.
Rationale: Maxima is an open-source software that can enable learners to solve linear
and quadratic equations, simultaneous equations, integration and differentiation,
perform algebraic manipulations: factorisation, simplification, expansion, etc This
resource is compulsory for learners taking distance learning as it enables them learn
faster using the ICT skills already learnt.
Resource #2 Graph
Complete reference : Copy of Graph on a disc is accompanying this course
Abstract : It is difficult to draw graphs of functions, especially complicated functions,
most especially functions in 3 dimensions. The learners, being distance learners, will
inevitably encounter situations that will need mathematical graphing. This course is
accompanied by a software called Graph to help learners in graphing. Learners
however need to familiarise with the Graph software to be able to use it.
Rationale: Graph is an open-source dynamic graphing software that learners can
access on the given CD. It helps all mathematics learners to graph what would
otherwise be a nightmare for them. It is simple to use once a learner invests time to
learn how to use it. Learners should take advantage of the Graph software because it
can assist the learners in graphing in other subjects during the course and after.
Learners will find it extremely useful when teaching mathematics at secondary school
level.
17
USEFUL LINKS
Useful Link #1
Title : Wikipedia
URL : http:www. http://en.wikipedia.org/wiki/Statistics
Screen capture :
Description: Wikipedia is every mathematician’s dictionary. It is an open-resource that
is frequently updated. Most learners will encounter problems of reference materials from
time to time. Most of the books available cover only parts or sections of Probability and
Statistics. This shortage of reference materials can be overcome through the use of
Wikipedia. It’s easy to access through “Google search”
Rationale: The availability of Wikipedia solves the problem of crucial learning materials
in all branches of mathematics. Learners should have first hand experience of
Wekipedia to help them in their learning. It is a very useful free resource that not only
solves student’s problems of reference materials but also directs learners to other
related useful websites by clicking on given icons. Its usefulness is unparalleled.
Useful Link #2
Title : Mathsguru
URL : http://en.wikipedia.org/wiki/Probability
Screen capture :
Description: Mathsguru is a website that helps learners to understand various
branches of number theory module. It is easy to access through Google search and
provides very detailed information on various probability questions. It offers
explanations and examples that learners can understand easily.
Rationale: Mathsguru gives alternative ways of accessing other subject related topics,
hints and solutions that can be quite handy to learners who encounter frustrations of
getting relevant books that help solve learners’ problems in Probability. It gives a helpful
approach in computation of probabilities by looking at the various branches of the
probability module.
18
Useful Link #3
Title : Mathworld Wolfram
URL : http:www. http://mathworld.wolfram.com/Probability
Screen capture :
Description: Mathworld Wolfram is a distinctive website full of Probability solutions.
Learners’ should access this website quite easily through Google search for easy
reference. Wolfram also leads learners to other useful websites that cover the same
topic to enhance the understanding of the learners.
Rationale:Wolfram is a useful site that provides insights in number theory while
providing new challenges and methodology in number theory. The site comes handy in
mathematics modelling and is highly recommended for learners who wish to study
number theory and other branches of mathematics. It gives aid in linking other webs
thereby furnishing learners with a vast amount of information that they need to
comprehend in Probability and Statistics.
19
1)
UNIT1: 40 HOURS)
DESCRIPTIVE STATISTICS AND PROBABILITY DISTRIBUTIONS
LEARNING ACTIVITY 1
A curious farmer undertakes the following activities in her farm.
1. Plants 80 tree seedlings on 1st March. She measures the heights of the
on 1st December.
2. She weighs all the 40 cows in her farm and records the weights in her
3. She records the daily production of eggs from the poultry section.
4. She records the time taken to deliver the milk to the processing plant.
The records are kept as below.
1.
2.
trees
diary.
Heights of plants in cm
77
76
62
85
63
68
82
67
75
68
74
85
71
53
78
60
81
80
88
73
75
53
95
71
85
74
73
62
75
61
71
68
69
83
95
94
87
78
82
66
60
83
60
68
77
75
75
78
89
96
72
71
76
63
62
78
61
65
67
79
75
53
62
85
93
88
97
79
73
65
93
85
76
76
90
72
57
84
73
86
Weights of goats in kg
Weight
(kg)
118126
127135
136144
145153
154162
163171
172180
No. of
goats
3
5
9
12
5
4
2
20
3.
4.
Number of laid eggs
Eggs
462
480
498
516
534
552
570
588
606
624
No of
days
98
75
56
42
30
21
15
11
6
2
Delivery time of milk to processing plant
Time
minutes
in
No. of days
90100
8089
7079
6069
5059
4049
3039
9
32
43
21
11
3
1
CASE 1:
A local firm dealing with agriculture extension services visits the farmer. She proudly
produces her records. The agricultural officer is very impressed by her good records
but clearly realises that the farmer needs some skills in data management to enable
her make informed decisions based on her farm outputs.
The agricultural officer designs a short course on data processing for all the rural
farmers.
During the course planning stage, the following terms are defined and designed for a
lesson one to the farmers.
a)
b)
c)
d)
e)
f)
Data : The result of observation e.g. height of tree seedlings
Frequency: Rate of occurrence e.g. number of goats weighed.
Mean: The average of a data
Mode: The highest occurring in a data.
Median: In an ascending data, the median is the term occurring at the middle of
the data.
Range: the difference between the highest and the lowest in the data.
LESSON ONE: MEASURES OF DISPERSION
Introduction to Statistics
Descriptive statistics is used to denote any of the many techniques used to
summarize a set of data. In a sense, we are using the data on members of a set to
describe the set. The techniques are commonly classified as:
1. Graphical description in which we use graphs to summarize data.
2. Tabular description in which we use tables to summarize data.
3. Parametric description in which we estimate the values of certain parameters
which we assume to complete the description of the set of data.
21
In general, statistical data can be described as a list of subjects or units and the data
associated with each of them. We have two objectives for our summary:
1. We want to choose a statistic that shows how different units seem similar.
Statistical textbooks call the solution to this objective, a measure of central
tendency.
2. We want to choose another statistic that shows how they differ. This kind of
statistic is often called a measure of statistical variability.
When we are summarizing a quantity like length or weight or age, it is common to
answer the first question with the arithmetic mean, the median, or the mode.
Sometimes, we choose specific values from the cumulative distribution function called
quartiles.
The most common measures of variability for quantitative data are the variance; its
square root, the standard deviation; the statistical range; interquartile range; and the
absolute deviation.
FARMERS LESSONS
The farmers are taught how to compute the
a)
Mean or Average of a data as follows:
Average of a data= Sum total of the data divided by number of items in data.
Example:
Calculate the mean of the following data:
1)
1,3,4,4,5,6,3,7,
Solution: Mean =
2)
1 3 4 4 5 6 3 7
33
=
= 4.125
8
8
650,675, 700, 725, 800, 900, 1050, 1125, 1200, 575
Solution:
650 675 700 725 800 900 1050 1125 1200 575
10
8400
=
10
= 840
Mean =
22
LESSON TWO
MEAN OF DISCRETE DATA
Example:
1) Find the mean of the following data:
X 22 24 25 33 36 37 41
f
5
7
8
4
6
9
11
Solution:
Mean
22(5) 24(7) 25(8) 33(4) 36(6) 37(9) 41(11) 1628
=
= 32.56
5 7 8 4 6 9 11
50
2) Find the mean wage of the workers:
Wage in $
220 250 300 350 375
No. of Workers 12
15
18
20
5
Solution:
220(12) 250(15) 300(18) 350(20) 375(5)
20665
Mean =
=
12 15 18 20 5
70
= $ 295.214
FREQUENCY TABLES AND MEAN OF GROUPED DATA
Example:
The weights of milk deliveries to a processing plant are shown below:
45 49 50 46 48 42 39 47 42 51
48 45 45 41 46 37 46 47 43 33
56 36 42 39 52 46 43 51 46 54
39 47 46 45 35 44 45 46 40 47
23
a) Using class intervals of 5, tabulate this data in a frequency table
b) Calculate the mean mass of the milk delivered.
Solution:
Frequency / Tally table
Class
Tally
Frequency
33- 37
////
4
37-42
///// ///
8
43-47
//// //// //// ///
19
48-52
//// //
7
53-57
//
2
Total
40
c) Mean of a grouped data
Class
Tally
Frequency(f)
Mid-point (x)
fx
33- 37
////
4
33 37
35
2
4 35 = 140
37-42
///// ///
8
40
320
43-47
//// //// //// ///
19
45
855
48-52
//// //
7
50
350
53-57
//
2
55
110
Total
40
Mean =
1775
fx 1775 44.375
f 40
24
DO THIS
Work out the mean of;
1).
63, 65, 67, 68, 69
2).
x
1
2
3 4 5
f(x) 11 10 5 3 1
3).
Weight (x)
4-8 9-13 14-18 19-23 24-28 29-33
Frequency 2
4).
4
7
14
8
5
91,78, 82,73,84
5).
Height (x)
61 64 67 70 73
Frequency 5
18 42 27 8
6).
Weight (x)
Frequency
Answer Key:
1).
66.4
4)
80
2).
5)
30.5-36.5 36.5-42.5 42.5-48.5 48.5-54.5 54.5-60.5
4
2.1
76.45
10
3).
6)
14
27
45
20.6
51.44
25
LESSON THREE
MODE
Example
1) Find the mode of the following data: 1,3,4,4,5,6,1,3,3,2,2,3,3,5
Solution:
The mode of a data is the item that appears most times. In this data, 3 occurs most
times or most frequently i.e. 5 times. Therefore the mode is 3.
2) Find the mode of the following data: 22, 24, 25,22, 27, 22, 25, 30, 25, 31
Solution:
22 and 25 occur three times each. Therefore the modes are 22 and 25. this is called a
bimodal data.
3) Find the mode of the data:
Observation
( X)
0
1
2
3
4
Frequency (
f)
3
7
10
16
11
Solution:
The most occurring observation is 3 i.e. 3 occurs 16 times.
4). Find the modal class of the following data
Weight ( X)
50 – 54 55-59 60-64 65-69 70-74 75-79 80-84
Frequency ( f) 3
6
8
5
15
9
13
Solution:
The modal class is 70-74 because it has the highest frequency of occurrence.
26
DO THIS
Work out the modes or modal classes of the following data;
1) 6, 8, 3,5,2,6,5,9,5
2) 20.4, 20.8, 22.1, 23.4, 19.7, 31.2, 23.4, 20.8, 25.5,23.4
3)
Weight (x)
4-8 9-13 14-18 19-23 24-28 29-33
Frequency 2
4
7
14
8
5
4)
Weight (x)
Frequency
Answer key:
1)
5
2)
30.5-36.5 36.5-42.5 42.5-48.5 48.5-54.5 54.5-60.5
4
23.4
10
3)
19-23
14
4)
27
45
54.5-60.5
LESSON FOUR
MEDIAN
The median is the value in the middle of a distribution e.g. in 1, 2,3,4,5, the median is 3
i.e it comes at exactly in the middle of the distribution. For the data 1,2,2,3,4,5,6,7,7,8;
there are 10 terms and no middle number. In such a case, the median is the average of
the two numbers bordering the centre line
Eg
1,2,2,3,
4
Therefore the median
5
6,7,7, 8
45
= 4.5
2
27
MEDIAN OF A GROUPED DATA
Example
Find the median of the following grouped data
Mass ( X)
20-24 25-29 30-34 35-39 40-44
Frequency (f) 4
10
16
8
2
Solution:
f
40 Therefore the median is the average of the 20th and 21st terms
20 21
= 10.5th term
2
Definition: Lower and Upper Limits of a Class.
The Lower Class Limit ( L.C.L) or lower class boundary and the Upper Class Limits
(U.C.L) or upper class boundary are the lower and upper bounds of a class interval e.g
the lower and upper limits of the class interval 20-24 are 19.5 and 20.5 and the L.C.L
and U.C.L of the class interval 35-39 are 34.5 and 39.5.
Mass ( X)
20-24 25-29
30- 34
35-39
40-44
16
8
2
Frequency (f) 4
10
Cumulative
4+10=14 14 + 16 = 30 30+8=38 39+2 =40
4
Frequency
Procedure for Calculation of the Median
Step 1:
Step 2:
Step 3:
Step 4:
The median occurs in the class interval 30-34
L.C.L and U.C.L of 30-34 are 29.5 and 34.5
Work out the Cumulative Frequency ( C.F)
Work out the class interval as U.C.L – L.C.L
Step 5: To get the 10.5th term.
28
10.5 th term L.C.L of class with median
Summation difference
Class Interval
Class frequency
i.e Summation difference 20.5 – 14 = 6.5 where 14 is the C.F of the class interval 25-29.
6.5
5 = 31.53125.
Step 6: The median = 29.5 +
16
Note that the denominator 16 is the class frequency in the class interval 30-34.
RANGE OF A DATA
The range of a data is simply the difference between the highest and the lowest score in
a data
Example: 23,26,34, 47,63 the range is 63-23=40 and in 121, 65, 78, 203, 298, 174 the
range is 298 – 65= 233.
LESSON FIVE: MEASURES OF DISPERSION
1)
QUARTILES
Data arranged in order of magnitude can be subdivided into four equal portions i.e. 25%
each. The first portion is the lower quartile occurring at 25%. The middle or centre
occurring at 50% is called the median while the third quarter occurring at 75% is called
the upper quartile. The three points are normally referenced as Q 1, Q2 , Q3
respectively.
2)
SEMI –INTERQUARTILE RANGE
The semi-interquartile range or the quartile deviation of a data is defined as
Q Q1
Q 3
2
3)
DECILES
If data arranged in order of magnitude is sub-divided into 10 equal portions ( 10% each),
then each portion constitutes a decile. The deciles are denoted by D1, D2, D3,……D9
4)
PERCENTILES
If data divided arranged in order of magnitude is subdivided into 100 equal portions
(1%each), then the portion constitutes a percentile. Percentiles are denoted as P 1, P2,
P3…, P99
29
THE MEAN DEVIATION
The mean deviation (average deviation), of a set of N numbers X1 ,X2, X3, X4, X5,……,
XN is defined by
N
Xj X
X X = X X , where X is the arithmetic
j 1
Mean deviation (MD) =
=
N
N
mean of the numbers and X X is the absolute value of the deviation of X
Example:
Find the mean deviation of the set
j
from X .
3, 4, 6, 8, 9.
Solution:
3 4 6 8 9 30
6
5
5
36 46 66 86 96
The mean deviation ( X ) =
=
5
Arithmetic mean =
3 2 0 2 3
5
=
3 2 0 2 3 10
5
5
2
THE MEAN DEVIATION OF A GROUPED DATA
For the data
Values
X1 X2 X3 …… XN
Frequencies f1
f2
f3
….
Fm
The mean deviation can be computed as
m
fj Xj X
f XX
j 1
Mean deviation =
XX
N
N
30
THE STANDARD DEVIATION
The Standard deviation of a set of N numbers X1 ,X2, X3, X4, X5,……, XN is denoted by s
and is defined by:
s=
N
2
(X j X )
j 1
N
=
( X X )2
=
N
x2
N
=
( X X )2
where x represents the deviations of the numbers X
from the mean X .
j
It follows that the standard deviation is the root mean square of the deviations
from the mean.
THE STANDARD DEVIATION OF A GROUPED DATA
Values
X1 X2 X3 …… XN
Frequencies f1
f2
f3
….
Fm
The standard deviation is calculated as:
s=
m
2
f j (X X )
j 1
N
2
f (X X )
N
2
fx
( X X )2
N
m
where N= f f .
j
j 1
THE VARIANCE
The variance of a set of data is defined as the square of the standard deviation i.e
variance = s2. We sometimes use s to denote the standard deviation of a sample of a
population and ( Greek letter sigma ) to denote the standard deviation of a
population population. Thus 2 can represent the variance of a population and s2 the
variance of sample of a population.
31
EXAMPLES
1)
Find the Mean and Range of the following data: 5,5,4,4,4,2,2,2
Solutions
n
8) Mean =
N
x =
55 4 4 4 4 2 2 2
356
.
9
n
or
0
0
9
=
52 44 23
9
= 3.56
9) Range 5 – 2 =3.
MEDIAN (MIDDLE )OBSERVATION
Example
Given 13 observations
1,1,2,3,4,4,5,6,8,10,14,15,17
The median falls =
The value
n 1 14
607
2
2
14
= 7th position. The median is 5
2
If n is odd the Median is the value in position
n 1
2
But if it is even, we consider the average of the two middle terms.
10) Example
1,1,2,2,3,4,4,5,6,8,10,14,15,17
The median = Average of the Middle two terms
45
4.5
=
2
32
Median of Grouped Data
When data are grouped the median 2 is the value at or below 50% of the observation
fall.
DO THIS
Find the median of the following data
1.
1,1,2,2,3,4,5,7,7,7,9
2.
7,8,1,1,9,19,11,2,3,4,8
Group Work
1. Study the computation of the variance and standard
from the following example.
Definition
The mean squared deviation from the mean is called variance:
h ( x x) 2
s2
N
Where: x x is deviation from the mean, N is number of observations
s 2 is variance and
s 2 is standard deviation.
Example
Given the data 2,4,5,8,11. Find the variance and the standard deviation.
33
X
2
4
5
8
11
x =5
xx
-4
-2
-1
2
5
( x x) 2
16
4
1
4
25
( x x) 2 =50
30
50
6
52
10
5
5
50
10
Variance= s 2
5
Standard deviation = √10.
So
x
DO THIS
1) Calculate range of the data:
1,1,1,2,2,3,3,3,4,5
10) Calculate the variance and the standard deviation: 1,2,3,4,5
SKEWNESS
Definition: Skewness is the degree of departure from symmetry of a distribution. ( Check
positive and negative skewness above)
For skewed distributions, the mean tends to lie on the same side of the mode as the
longer tail.
PEARSON’S FIRST COEFFICIENT OF SKEWNESS
This coefficient is defined as
mean mod e
X mod e
Skewness=
s tan dard deviation
s
34
PEARSON’S SECOND COEFFICIENT OF SKEWNESS
This coefficient is defined as:
Skewness=
3(mean median)
3( X median)
s tan dard deviation
s
QUARTILE COEFFICIENT OF SKEWNESS
This is defined as:
Quartile coefficient of skewness =
(Q Q ) (Q Q ) Q 2Q Q
3
2
2
1 3
2
1
Q Q
Q Q
3
1
3
1
10-90 PERCENTILE OF SKEWNESS
This is defined as:
(P P ) (P P ) P 2P P
50
50 10 90
50 10
10-90 percentile of skewness = 90
P P
P P
90 10
90 10
Example: Find 25th percentile of the data 1, 2, 3, 4, 5, 6, 7, 9
25th percentile =
(n 1) x0.25
9(.25) 22.5( percentile )
2nd = 2
3rd = 3
2.25 0.25(1) 2 2.25
Find 50th percentile
50th percentile: (8 1) x.50 9(.5) 4.5 percentile
4th = 4
5th = 5
0.5(5) 0.5 4 4.5
The (1) is the range 5 4 1
Group Work
1. Study the computation of percentiles and attempt
the following question..
35
DO THIS
Find the 25th percentile, the 50th percentile, and 90th percentile
46,21,89,42,35,36,67,53,42,75,42,75,47,85,40,73,48,32,41,20,75,48,48,32,52,61,49,50,
69,59,30,40,31,25,43,52,62,50
Answer Key
a)
36
b)
48
c)
73
KURTOSIS
Definition: Kurtosis is the degree of peakedness of a distribution, as compared to the
normal distribution.
EXAMPLES:
1)
LEPTOKURTIC DISTRIBUTION
A distribution having a relatively high peak
2)
PLATYKURTIC DISTRIBUTION
A distribution having a relatively flat top
3). MESOKURTIC DISTRIBUTION
A Normal Distribution – not very peaked or flat topped
36
DO THIS
Find the mode for the data collection:
1)
1,3,4,4,2,3,5,1,3,3,5,4,2,2,2,3,3,4,4,5
2)
Number of marriage per 1000 persons in Africa population for years 1965 – 1975
Year
1965
1966
1967
1968
1969
1970
1971
1972
1973
1974
1975
3)
Rate
9.3
9.5
9.7
10.4
10.6
10.6
10.6
10.9
10.8
10.5
10.0
Number of deaths per 1000 years for years 1960 and 1965 – 1975
1960
1965
1966
1967
1968
1969
1970
1971
1972
1973
1974
1975
9.5
9.4
9.5
9.4
9.7
9.5
9.5
9.3
9.4
9.3
9.1
8.8
SOLUTIONS
1.
3
2.
10.6
3.
9.5
37
READ:
1) An Introduction to Probability by Charles M.
Grinstead pages 247 -263
Exercise on pg 263-267 Nos. 4,7,8,9
PROBABILITY
1) Sample Space and Events
Terminology
a) A Probability experiment
When you toss a coin or pick a card from a deck of playing cards or roll a dice, the act
constitutes a probability experiment. In a probability experiment, the chances are well
defined with equal chances of occurrence e.g. there are only two possible chances of
occurrence in tossing a coin. You either get a head or tail. The head and the tail have
equal chances of occurrence.
b) An Outcome
This is defined as the result of a single trial of a probability experiment e.g. When you
toss a coin once, you either get head or tail.
c). A trial
This refers to an activity of carrying out an experiment like picking a card from a deck of
cards or rolling a die or dices.
d). Sample Space
This refers to all possible outcomes of a probability experiment. e.g. in tossing a coin,
the outcomes are either Head(H) or tail(T) i.e there are only two possible outcomes in
tossing a coin. The chances of obtaining a head or a tail are equal.
e). A Simple and Compound Events
In an experimental probability, an event with only one outcome is called a simple event.
If an event has two or more outcomes, it is called a compound event.
2) Definition of Probability.
Probability can be defined as the mathematics of chance. There are mainly four
approaches to probability;
1) The classical or priori approach
2) The relative frequency or empirical approach
3) The axiomatic approach
4) The personalistic approach
The Classical or Priori Approach
Probability is the ratio of the number of favourable cases as compared to the total likely
cases. Suppose an event can occur in N ways out of a total of M possible ways. Then
the probability of occurrence of the event is denoted by
38
N
. Probability refers to the ratio of possible outcomes to all possible
M
outcomes.
The probability of non-occurrence of the same event is given by {1-p(occurrence)}. The
probability of occurrence plus non-occurrence is equal to one.
If probability occurrence; p(O) and probability of non-occurrence (O’), then
p(O)+p(O’)=1.
p=Pr(N)=
Empirical Probability ( Relative Frequency Probability)
Empirical probability arises when frequency distributions are used.
For example:
Observation ( X)
0
1
2
3
4
Frequency ( f)
3
7
10
16
11
The probability of observation (X) occurring 2 times is given by the formulae
P(2)=
freuency of 2
f (2)
10
10
sum of frequencies f 3 7 10 16 11 47
3) Properties of Probability
a) Probability of any event lies between 0 and 1 i.e. 0 p(O) 1. It follows that
probability cannot be negative nor greater than 1.
b) Probability of an impossible event ( an event that cannot occur ) is always zero(0)
c) Probability of an event that will certainly occur is 1.
d) The total sum of probabilities of all the possible outcomes in a sample space is
always equal to one(1).
e) If the probability of occurrence is p(o)= A, then the probability of non-occurrence
is 1-A.
COUNTING RULES
1) FACTORIALS
Definition: Factorial 4 ! = 4 x 3 x 2 x 1 and 7! = 7 x 6 x 5 x 4 x 3 x 2 x 1
39
2) PERMUTATION RULES
n !
Definition: nPr =
(n r ) !
Examples:
5!
5 x 4 x3x 2 x1
5P3 =
5 x 4 x3 60
(5 3)!
2 x1
8!
8! 8 x7 x6 x5 x 4 x3x 2 x1
8P5 =
8 x7 x6 x5 x 4 6720
(8 5)! 3!
3x 2 x1
3) COMBINATIONS
Definition: nCr =
n !
(n r ) ! r !
Examples:
5!
5 x 4 x3x 2 x1 5 x 4
10
(5 2)!2!
3! 2!
2 x1
10 x9 x8 x7 x 6! 10 x9 x8 x7
10!
10!
210
10C6 =
(10 6)!6! 4! 6!
4 x3x 21x 6!
4 x3x 2 x1
5C2 =
DO THIS
Work out the following;
1).
8P3
2)
8C3
3)
15C10
4)
6C3
5)
15P4
6)
9C3
7)
10C8
8)
7P4
Answer key
1)
5)
336
32 760
2)
6)
56
84
3)
7)
3003
90
4)
8)
20
840
40
RULES OF PROBABILITY
ADDITION RULES
1) Rule 1: When two events A and B are mutually exclusive, then
P(A or B)=P(A)+P(B)
Example: When a is tossed, find the probability of getting a 3 or 5.
Solution: P(3) =1/6 and P(5) =1/6.
Therefore P( 3 or 5) = P(3) + P(5) = 1/6+1/6 =2/6=1/3.
2) Rule 2: If A and B are two events that are NOT mutually exclusive, then
P(A or B) = P(A) + P(B) - P(A and B), where A and B means the number of
outcomes that event A and B have in common.
Example: When a card is drawn from a pack of 52 cards, find the probability that the
card is a 10 or a heart.
Solution:
P( 10) = 4/52 and P( heart)=13/52
P ( 10 that is Heart) = 1/52
P( A or B) = P(A) +P(B)-P( A and B) = 4/52 _ 13/52 – 1/52 = 16/52.
MULTIPLICATION RULES
1) Rule 1: For two independent events A and B, then P( A and B) = P(A) x P(B).
Example: Determine the probability of obtaining a 5 on a die and a tail on a coin in
one throw.
Solution: P( 5) =1/6 and P(T) =1/2.
P(5 and T)= P( 5) x P(T) = 1/6 x ½= 1/12.
2) Rule 2: When to events are dependent, the probability of both events occurring is
P(A and B)=P(A) x P(B|A), where P(B|A) is the probability that event B occurs given that
event A has already occurred.
Example: Find the probability of obtaining two Aces from a pack of 52 cards without
replacement.
Solution: P( Ace) =2/52 and P( second Ace if NO replacement) = 3/51
Therefore P(Ace and Ace) = P(Ace) x P( Second Ace) = 4/52 x 3/51 = 1/221
CONDITIONAL PROBABILITY
P( A and B)
, where
P( B)
P(A and B) means the probability of the outcomes that events A and B have in common.
Example: When a die is rolled once, find the probability of getting a 4 given that an
even number occurred in an earlier throw.
Solution: P( 4 and an even number) = 1/6 ie. P(A and B) =1/6. P(even number) =3/6
=1/2.
P( A and B) 16 1
P( A|B) =
1
P( B)
3
2
The conditional probability of two events A and B is P(A|B) =
41
EXAMPLES:
1) A bag contains 3 orange, 3 yellow and 2 white marbles. Three marbles are selected
without replacement. Find the probability of selecting two yellow and a white marble.
Solution. P( 1st Y) =3/8, P( 2nd Y) = 2/7 and P( W)= 2/6
P(Y and Y and W)=P(Y) x P(Y) x P(W) = 3/8 x 2/7 x 2/6 = 1 / 28
2) In a class, there are 8 girls and 6 boys. If three students are selected at random for
debating, find the probability that all girls.
Solution: P( G) =8/14 and P(B) =6/14. P( 1st G)=8/14, P(2nd G) 7/13 and
6/12.
P( three girls) 8/14 x 7/13 x 6/12= 2/13
P(3rdG)=
3) In how many ways can 3 drama officials be selected from 8 members?
Solution: 8C3 = 56 ways.
4) A box has 12 bulbs, of which 3 are defective. If 4 bulbs are sold, find the probability
that exactly one will be defective.
Solution:
P( defective bulb)= 3C1 and P( non-defective bulbs) = 9C3
3!
9!
x
252
3C1 x 9C3 =
(3 1)!1! (9 3)!3!
P( 4 bulbs from 12) = 12C4 = 495.
P( 1 defective bulb and 3 okey bulbs) = 295/495=0.509.
DO THIS
1) In how many ways can 7 dresses be displayed in a row on a shelf?
2) In how many ways can 3 pens be selected from 12 pens?
3) From a pack of 52 cards, 3 cards are selected. What is the probability that they will
all be diamonds?
Answer Key:
1). 5040
2). 220
3). 0.013
42
READ :An Introduction to Probability & Random Processes
By Kenneth B & Gian-Carlo R, pages
1. 1.20 -1.22
Exercise Chapter 1: Sets, Events & Probability Pg
1.23-1.28 Nos. 1-12 & 14-20
2. 2.1-2.33
Exercise Chapter 2: Finite Processes Pg 2.33 Nos.
1,2,3,13-20, 22-27
3. Introduction to Probability, By Charles M. Grinstead
pages139-141
RANDOM VARIABLES
Random Variables ( r.v)
Definition: A random variable is a function that assigns a real number to
every possible result of a random experiment.
(Harry Frank & Steve C Althoen,CUP, 1994, pg 155)
A random variable is a variable in the sense that it can be used as a placeholder for a
number in equations and inequalities. Its randomness is completely described by its
cumulative distribution function which can be used to determine the probability it takes
on particular values.
Formally, a random variable is a measurable function from a probability space to the
real numbers. For example, a random variable can be used to describe the process of
rolling a fair die and the possible outcomes { 1, 2, 3, 4, 5, 6 }. The most obvious
representation is to take this set as the sample space, the probability measure to be
uniform measure, and the function to be the identity function.
Random variable
Some consider the expression random variable a misnomer, as a random variable is not
a variable but rather a function that maps outcomes (of an experiment) to numbers. Let
A be a σ-algebra and Ω the space of outcomes relevant to the experiment being
performed. In the die-rolling example, the space of outcomes is the set Ω = { 1, 2, 3, 4,
5, 6 }, and A would be the power set of Ω. In this case, an appropriate random variable
might be the identity function X(ω) = ω, such that if the outcome is a '1', then the
random variable is also equal to 1. An equally simple but less trivial example is one in
which we might toss a coin: a suitable space of possible outcomes is Ω = { H, T } (for
heads and tails), and A equal again to the power set of Ω. One among the many
possible random variables defined on this space is
43
Mathematically, a random variable is defined as a measurable function from a sample
space to some measurable space.
CONVERGENCE OF RANDOM VARIABLES
In probability theory, there are several notions of convergence for random variables.
They are listed below in the order of strength, i.e., any subsequent notion convergence
in the list implies convergence according to all of the preceding notions.
Convergence in distribution: As the name implies, a sequence of random variables
converges to the random variable
in distribution if their respective
cumulative distribution functions
function of
, wherever is continuous.
converge to the cumulative distribution
Weak convergence: The sequence of random variables
is said to
converge towards the random variable
weakly if
every ε > 0. Weak convergence is also called convergence in probability.
Strong convergence: The sequence of random variables
converge towards the random variable
strongly if
convergence is also known as almost sure convergence.
for
is said to
Strong
Intuitively, strong convergence is a stronger version of the weak convergence, and in
both cases the random variables
show an increasing correlation with
However, in case of convergence in distribution, the realized values of the random
variables do not need to converge, and any possible correlation among them is
immaterial.
.
Law of Large Numbers
If a fair coin is tossed, we know that roughly half of the time it will turn up heads, and the
other half it will turn up tails. It also seems that the more we toss it, the more likely it is
that the ratio of heads:tails will approach 1:1. Modern probability allows us to formally
arrive at the same result, dubbed the law of large numbers. This result is remarkable
because it was nowhere assumed while building the theory and is completely an
offshoot of the theory. Linking theoretically-derived probabilities to their actual frequency
of occurrence in the real world, this result is considered as a pillar in the history of
statistical theory.
44
The strong law of large numbers (SLLN) states that if an event of probability p is
observed repeatedly during independent experiments, the ratio of the observed
frequency of that event to the total number of repetitions converges towards p strongly
in probability.
In other words, if
are independent Bernoulli random variables taking values 1
with probability p and 0 with probability 1-p, then the sequence of random numbers
converges to p almost surely, i.e.
CENTRAL LIMIT THEOREM
The central limit theorem is the reason for the ubiquitous occurrence of the normal
distribution in nature, for which it is one of the most celebrated theorems in probability
and statistics.
The theorem states that the average of many independent and identically distributed
random variables tends towards a normal distribution irrespective of which distribution
the original random variables follow. Formally, let
variables with means
random variables
, and variances
be independent random
Then the sequence of
converges in distribution to a standard normal random variable.
FUNCTIONS OF RANDOM VARIABLES
If we have a random variable X on Ω and a measurable function f: R → R, then Y = f(X)
will also be a random variable on Ω, since the composition of measurable functions is
also measurable. The same procedure that allowed one to go from a probability space
(Ω, P) to (R, dFX) can be used to obtain the distribution of Y. The cumulative distribution
function of Y is
45
Example
Let X be a real-valued, continuous random variable and let Y = X2. Then,
If y < 0, then P(X2 ≤ y) = 0, so
If y ≥ 0, then
So
PROBABILITY DISTRIBUTIONS
Certain random variables occur very often in probability theory due to many natural and
physical processes. Their distributions therefore have gained special importance in
probability theory. Some fundamental discrete distributions are the discrete uniform,
Bernoulli, binomial, negative binomial, Poisson and geometric distributions. Important
continuous distributions include the continuous uniform, normal, exponential, gamma
and beta distributions.
DISTRIBUTION FUNCTIONS
If a random variable
defined on the probability space (Ω,A,P) is given, we
can ask questions like "How likely is it that the value of X is bigger than 2?". This is the
same as the probability of the event
P(X > 2) for short.
which is often written as
Recording all these probabilities of output ranges of a real-valued random variable X
yields the probability distribution of X. The probability distribution "forgets" about the
particular probability space used to define X and only records the probabilities of various
values of X. Such a probability distribution can always be captured by its cumulative
distribution function
and sometimes also using a probability density function. In measure-theoretic terms, we
use the random variable X to "push-forward" the measure P on Ω to a measure dF on
R. The underlying probability space Ω is a technical device used to guarantee the
46
existence of random variables, and sometimes to construct them. In practice, one often
disposes of the space Ω altogether and just puts a measure on R that assigns measure
1 to the whole real line, i.e., one works with probability distributions instead of random
variables.
DISCRETE PROBABILITY THEORY
Discrete probability theory deals with events which occur in countable sample
spaces.
Examples: Throwing dice, experiments with decks of cards, and random walk.
Classical definition: Initially the probability of an event to occur was defined as number
of cases favorable for the event, over the number of total outcomes possible.
For example, if the event is "occurrence of an even number when a die is rolled", the
probability is given by
, since 3 faces out of the 6 have even numbers.
Modern definition: The modern definition starts with a set called the sample space
which relates to the set of all possible outcomes in classical sense, denoted by
. It is then assumed that for each element
, an intrinsic
"probability" value
is attached, which satisfies the following properties:
1.
2.
An event is defined as any subset
event defined as
of the sample space
. The probability of the
So, the probability of the entire sample space is 1, and the probability of the null event is
0.
The function
mapping a point in the sample space to the "probability" value is
called a probability mass function abbreviated as pmf. The modern definition does
not try to answer how probability mass functions are obtained; instead it builds a theory
that assumes their existence.
47
CONTINUOUS PROBABILITY THEORY
Continuous probability theory deals with events which occur in a continuous sample
space.
If the sample space is the real numbers, then a function called the cumulative
distribution function or cdf
is assumed to exist, which gives
.
The cdf must satisfy the following properties.
1.
is a monotonically non-decreasing right-continuous function
2.
3.
If
is differentiable, then the random variable is said to have a probability density
function or pdf or simply density
For a set
.
, the probability of the random variable being in
is defined as
In case the density exists, then it can be written as
Whereas the pdf exists only for continuous random variables, the cdf exists for all
random variables (including discrete random variables) that take values on .
These concepts can be generalized for multidimensional cases on
.
48
PROBABILITY DENSITY FUNCTION
DISCRETE DISTRIBUTION
If X is a variable that can assume a discrete set of values X1, X2, X3,…….., Xk wih respet
to probabilities p1, p2, p3,……., pk, where p1+ p2 + p3,……., + pk = 1, we say that a
discrete probability distribution for X has been defined. The function p(X), which has the
respective values p1, p2, p3,……., pk for X= X1, X2, X3,…….., Xk is called the probability
function, or frequency function, of X. Because X can assume certain values with given
probabilities, it is often called a discrete random variable. A random variable is also
known as a chance variable or stochastic variable. { Murray R, 2006 pg 130}
CONTINUOUS DISTRIBUTION
Suppose X is a continuous random variable. A continuous random variable X is
specified by its probability density function which is written f(x) where f(x) 0 throughout
the range of values for which x is valid. This probability density function can be
represented by a curve, and the probabilities are given by the area under the curve.
P(X)
X
a
b
The total area under the curve is equal to 1. The are under the curve between the lines
x=a and x=b ( shaded) gives the probability that X lies between a and b, which can be
denoted by P(a<X<b). p(X) is called a probability density function and the variable X is
often called a continuous random variable
Since the total area under the curve is equal to 1, it follows that the probability between
a range space a and b is given by
P ( a X b)
b
f ( x)dx ,
a
which is the shaded area.
49
Note: when computing area from a to b, we need not distinguish
( and ) and ( and ) inequalities. We assume the lines at a and b have no
thickness and its area is zero.
Solved Examples:
1) The continuous random variable X is distributed with probability density function f
defined by
f(x) = kx(16-x2), for 0<x<4.
Evaluate:
a). The value of constant k
b). The probability of range space P(1<X<2)
c). The probability P(x 3)
Solution:
f(x)
x
a
b
For any function f(x) such tha
f(x) 0, for a x b,
and
b
a f ( x)dx 1
may be taken as the probability density function (p.d.f) of a continuous random variable
in the range space a x b.
50
Procedure:
Step 1:
In general, if X is a continuous random variable (r.v) with p.d.f f(x) valid
over the range a x b, then
f ( x)dx 1
all x
b
a
i.e.
f ( x)dx 1
Step 2:
a). To determine k, we use the fact that in f(x) = kx(16-x2), for 0<x<4, then
4
0
kx(16 x 2 )dx 1
4
k 16 x x 3 ) dx 1
k
0
1
64
Step 3:
b).
Find P(1<X<2)
Solution:
2
P(1<X<2)= 1
f ( x)dx
1 2
81
3
(
16
x
x
)
dx
64 1
256
Step 4:
c). To find P(x 3)
51
P ( x 3)
1 4
49
3
(
16
x
x
)
dx
64 3
256
Example 2:
2).
X is the continuous random variable ‘the mass of a substance, in kg, per
minute in an industrial production process’, where
1
x (6 x )
f ( x) 12
0
(0 x 3)
otherwise
Find the probability that the mass is more than 2 kg.
Solution:
X can take values from 0 to 3 only. We sketch f(x), and shade the area
required.
f ( x)
f(x)
1
x (6 x )
12
x
0
2
3
1
x(6 x)dx
2 12
P ( x 2)
1
12
3
2
3
(6 x x 2 ) dx
3
x3
2
3 x
3 2
0.722 (3 d . p )
1
12
The probability that the mass is more than 2 kg is 0.722.
Worked example:
3).
A continuous random variable has p.d.f f(x) where
52
f ( x) kx2 , 0 x 6.
a).
Find the value of k
b).
Find P (2 X 4)
Solution:
a). Since X is a random variable the total probability is 1. i.e.
f ( x ) dx 1
all
6
0
k x2 dx 1
6
k x3
1
3
0
216 k
1
3
3
k
216
Therefore f(x)=
3 2
1 2
x
x , 0 x6
216
72
b).
f ( x)
f(x)
1
x2
72
x
0
2
4
P ( 2 x 4)
1
x3
216
0.259
6
4
2
1
x 2 dx
72
42
Therefore the probability P (2 X 4) = 0.259
53
Worked Example:
4).
The continuous random variable (r.v) has a probability density function(p.d.f)
where
0 x2
k
f ( x) k (2 x 3)
2 x5
0
otherwise
a).
Find the value of the constant k
b).
Sketch y=f(x)
c).
Find P(X 1)
d).
Find P(X>2.5)
Solution:
a).
Since X is a r.v, then
f ( x)dx 1
all x
Therefore
2
0
kx
5
kdx k (2 x 3)dx 1
2
2
0
k x 2 3x
5
2
2k 19k 1
1
k
21
b).
So the p.d.f of X is
1
21
1
( 2 x 3)
f ( x ) 21
0
0 x 2
2 x 5
otherwise
54
SKETCH
1
3
1
21
0
1
2
2.5
3
4
5
1
1
= = 0.048
21 21
c).
P(x 1) = area between zero and 1 = L x W= 1 x
d).
Find P(X>2.5) = area of rectangle + area of trapezium.
=(
1
1
1
2
11
0.131
x 2 ) + ( {0.5}{ } =
2
21 21 84
21
55
REFLECTION: It is important for Mathematics teachers to learn how to use
ICT in Graphs. The link opens up avenue for Maths teachers to learn how
to draw graphs using free software
http://www.chartwellyorke.com/
56
DO THIS
1).
2).
3).
The continuous random variable X has p.d.f f(x) where f(x)= k, 0 x 3 .
a)
Sketch y=f(x)
b).
Find the value of the constant k
c).
Find P(0.5 X 1
The continuous random variable has p.d.f f(x) where f(x)=kx2, 1 x 4 .
a).
Find the value of the constant
b).
Find P(x 2)
c).
Find P(2.5 x 3.5
The continuous random variable has p.d.f f(x) where
k
f ( x) k (2 x 1)
0
a)
b)
c)
d)
0 x2
2 x3
otherwise
find the value of the constant k.
Sketch y=f(x)
Find P(X 2 )
Find P(1 X 2.2)
EXPECTATION
Definition:
If X is a continuous variable (r.v) with probability density function (p.d.f) f(x), then the
expectation of X is E(X) where
E( X )
x
f ( x)dx
all x
NB: E(X) is often denoted by
and referred to as the mean of X
57
Example:
1).
If X is a continuous variable ( r.v) with a p.d.f
f ( x)
1 2
x , 0 x 3,
16
find E(X).
Solution:
E( X )
x
f ( x)dx
all x
3
0
1
{x} x 2 dx
16
3
1 x4
81
1.265
16 4 0
64
2).
If the continuous random variable X has p.d.f
f ( x)
2
(3 x)( x 1),
5
1 x 3,
find E(X).
E( X )
x
f ( x)dx
all x
E ( x)
3
1
2
{x} (3 x)( x 1)dx
5
3
2 x 4 2 x 3 3x 2
5 4
3
2 1
608
60
10.13
58
GENERALISATION:
If g( x) is any function of the continuous random variable r.v X having p.d.f f(x), then
E g ( X )
g ( x) f ( x)dx
all x
and in particular
E( X 2 )
x f x dx
2
all x
The following conclusion s hold
1.
E (a) a
2.
3.
E ( aX ) aE ( X )
E ( aX b) aE ( X ) b
4.
E ( f1 ( X ) f 2 ( X ) E f 2 ( X )
Example:
1).
The continuous random variable X has p.d.f f(x) where f(x)=
1
x, 0 x 3.
2
Find
a).
b).
c).
E(X)
E(X2)
E(2X +3)
Solution:
a)
E( X )
x
f ( x)dx
all x
3
0
1 2
x dx
2
3
1 x3
2
3 0
4 .5
59
b)
E( X
1
2
2
3
0
)
x
2
all x
f ( x ) dx
x 3 dx
3
1 x4
2 4 0
81
10.125
8
c).
E(2X +3) = E (2X) + 3
= 2E(X) +3
= 2(10.125)+5
= 25.25 ( from (b) above)
DO THIS
1).
The continuous random variable X has p.d.f f(x) where
kx
f ( x) k
k ( 4 x )
0
0 x 1
1 x 3
3 x5
otherwise
a).
b)
Find k
Calculate E(X)
2).
The continuous random variable has p.d.f f(x) where f(x) =
a).
b).
c).
d).
Find E(X)
Find E(2X+4)
Find E(X2).
Find E( X2 + 2X – 1).
1
( x 3), 0 x 5
10
60
BERNOULLI DISTRIBUTION
In probability theory and statistics, the Bernoulli distribution, named after Swiss
scientist Jakob Bernoulli, is a discrete probability distribution, which takes value 1 with
success probability p and value 0 with failure probability q = 1 − p. So if X is a random
variable with this distribution, we have:
The probability mass function f of this distribution is
The expected value of a Bernoulli random variable X is
, and its variance is
The kurtosis goes to infinity for high and low values of p, but for p = 1 / 2 the Bernoulli
distribution has a lower kurtosis than any other probability distribution, namely -2.
The Bernoulli distribution is a member of the exponential family.
BINOMIAL DISTRIBUTION
In probability theory and statistics, the binomial distribution is the discrete probability
distribution of the number of successes in a sequence of n independent yes/no
experiments, each of which yields success with probability p. Such a success/failure
experiment is also called a Bernoulli experiment or Bernoulli trial. In fact, when n = 1,
the binomial distribution is a Bernoulli distribution. The binomial distribution is the basis
for the popular binomial test of statistical significance.
Examples
An elementary example is this: roll a die ten times and count the number of 1s as
outcome. Then this random number follows a binomial distribution with n = 10 and p =
1/6.
For example, assume 5% of the population is green-eyed. You pick 500 people
randomly. The number of green-eyed people you pick is a random variable X which
follows a binomial distribution with n = 500 and p = 0.05 (when picking the people with
replacement).
61
EXAMPLES:
1) A coin is tossed 3 times. Find the probability of getting 2 heads and a tail in any
given order.
FORMULA:
We can use the formula
nCx.
(p)x.(1-p)n-x
Where n = the total number of trials
x = the number of successes ( 1,2,…)
p= the probability of a success.
1st) nCx determines the number of ways a success can occur.
2nd) (p)x is the probability of getting x successes and
3rd) (1-p)n-x is the probability of getting n-x failures
Solution:
Tossing 3 times means n=3
Two heads means x=2
P(H)=1/2; P(T)=1/2
1
1
P( 2 heads) = 3C2. ( )2.(1- )3-1 = 3(1/4)(1/2)= 3/8
2
2
DO THIS
1) Find the probability of exactly one 5 when a die is rolled 3 times
2) Find the probability of getting 3 heads when 8 coins are tossed.
3) A bag contains 4 red and 2 green balls. A ball is drawn and replaced 4 times. What
is the probability of getting exactly 3 red balls and 1 green ball.
Answer:
1
5
1).
P( one 5)
= 3C1. ( )1.( )2 = 25/72 = 0.347 i.e n=3, x=1, p=1/6
6
6
1
1
2).
P ( 3 heads) = 8C3. ( )3.( )5 = 7/32 = 0.218. i.e n=8, x=3, p=1/2
2
2
2
1
3).
P( 3 Red balls) = 4C3. ( )3.( )1 = 32/81= 0.395 i.e. n=4, x=3, p=2/3
3
3
62
READ:
1. Lectures on Statistics, By Robert B. Ash, , page 1-4
Exercise Nos.1, 2 and 3 on pg 4.
2. An Introduction to Probability & Random Processes By
Kenneth B & Gian-Carlo R, pages 3.1-3.63
Exercise Chapter 3: Random Variables pg 3.64-3.82
Nos. 1-7, 11-17, 20-24, 34-36
3. An Introduction to Probability By Charles M. Grinstead
pages 96-107, & 184
Exercise on pages 113-118 Nos.
1,2,3,4,5,8,9,10,19,20
Ref: http://en.wikipedia.org/wiki/measurable_space
Ref: http://en.wikipedia.org/wiki/Probability_theory
Ref: http://en.wikipedia.org/wiki/Bernoulli_distribution
POISSON DISTRIBUTION
In probability theory and statistics, the Poisson distribution is a discrete probability
distribution that expresses the probability of a number of events occurring in a fixed
period of time if these events occur with a known average rate, and are independent of
the time since the last event.
The distribution was discovered by Siméon-Denis Poisson (1781–1840)
The Poisson distribution is sometimes called a Poissonian, analagous to the term
Gaussian for a Gauss or normal distribution.
The Poisson distribution is used when the variable occurs over a period of time, volume,
area etc…it can be used for the arrival of airplanes at airports, the number of phone
calls per hour for a station, the number of white blood cells on a certain area.
The probability of x successes is
e x
x!
where e is a mathematical constant = 2.7183
is the mean or expected value of the variables.
63
Group Work
1. Study the probability computations and attempt the
given question.
Example
If there are 100 typographical errors randomly distributed. In 500 pages manuscripts
find the probability that any given page has exactly 4 errors.
Solution
Find the mean number of errors x
100 1
0 .2
100 5
In other words there is an average of 0.2 errors per page. In this case 4 so the
probability of selecting a page with exactly 4 errors
e .x x 2.7183 0.2
x!
41
0.2
4
= 0.00168
Amount 0.2%
Worked Example
A hot line with a full free number receives an average of 4 calls per hour for any given
hour. Find the probability that it will receive exactly 5 calls.
e . x 2.7183
x!
5!
3
35
= 0.1001
Which is 10%
64
DO THIS
1) A telephone Marketing Company gets an average of 5 orders per 1000 calls. If a
company calls 500 people find the probability of getting 2 orders.
SOLUTION
0.26
Which is 26%
READ :
1. An Introduction to Probability & Random Processes By
Kenneth B & Gian-Carlo R, pages187-192
2. Robert B. Ash, Lectures on Statistics, page 1 and
Answer problems 1,2,3 on pg 15.
Ref: http://en.wikipedia.org/wiki/Normal_distribution
GEOMETRIC DISTRIBUTION
In probability theory and statistics, the geometric distribution is either of two discrete
probability distributions:
the probability distribution of the number X of Bernoulli trials needed to get one
success, supported on the set { 1, 2, 3, ...}, or
the probability distribution of the number Y = X − 1 of failures before the first
success, supported on the set { 0, 1, 2, 3, ... }.
Which of these one calls "the" geometric distribution is a matter of convention and
convenience.
65
If the probability of success on each trial is p1, then the probability that k trials are
needed to get one success is
for k = 1, 2, 3, ....
Equivalently, if the probability of success on each trial is p0, then the probability that
there are k failures before the first success is
for k = 0, 1, 2, 3, ....
In either case, the sequence of probabilities is a geometric sequence.
For example, suppose an ordinary die is thrown repeatedly until the first time a "1"
appears. The probability distribution of the number of times it is thrown is supported on
the infinite set { 1, 2, 3, ... } and is a geometric distribution with p1 = 1/6.
SOLUTIONS USING THE GEOMETRIC DISTRIBUTION FORMULA
The formula for the probability that the first success occurs on the nth trial is
(1-p)n-1p or simply
, where p is the probability of a
success and n is the trial number of the first success.
Example:
1) Find the probability that the first tail occurs on the third toss of a coin.
Solution:
The outcome of a tail on the
1
therefore P(HHT) = ( 1- )3-1
2
third throw implies HHT. From (1-p)n-1p , n=3, p=1/2 and
1
1
1 1
( ) = ( ) .. ( ) ( ) =1/8
2
2
2 2
EXAMPLES IN GEOMETRIC DISTRIBUTION
Flipping a coin several times we apply the geometric to distribution to get the answer of
flipping a coin several times.
Example
1) A coin is tossed find the probability that the first head occurs on the third toss
solution. Out come is TTH
n = 3 and p=1/2
66
Probability of getting 2 tails and then one head is
1 1 1 1
2 2 2 8
Or by the formula
1
1
2
31
2
1 1 1 1
. .
2 2 2 8
2) A die is rolled; find the probability of getting the first 3 on the fourth roll.
Solution:
n=4
p=1/6
4 1
3
3
1 1 5 5 1 125
1
0.96
6 6 6 6 6 1296
Example
If cards are selected from a deck and replaced, how many trials would it take on
average to get two clubs?
P (Club) = 13/52=1/4
Expected no. of trials for selecting 2 clubs would be
2
4
2x 8
1
1
4
67
DO THIS
1. A card from an ordinary deck of cards is selected and then replaced with another
card selected etc… find the probability that the first club will occur on he fourth
draw.
2. A die is tossed until 5 or 6 is obtained. Find the expected number of tosses.
Answer Key:
1) Fourth
2) 3
HYPERGEOMETRIC DISTRIBUTION
In probability theory and statistics, the hypergeometric distribution is a discrete
probability distribution that describes the number of successes in a sequence of n draws
from a finite population without replacement.
A typical example is illustrated by the contingency table above: there is a shipment of N
objects in which D are defective. The hypergeometric distribution describes the
probability that in a sample of n distinctive objects drawn from the shipment exactly k
objects are defective.
In general, if a random variable X follows the hypergeometric distribution with
parameters N, D and n, then the probability of getting exactly k successes is given by
The probability is positive when k is between max{ 0, D + n − N } and min{ n, D }.
The formula can be understood as follows: There are
replacement). There are
there are
possible samples (without
ways to obtain k defective objects and
ways to fill out the rest of the sample with non-defective objects.
68
When the population size is large compared to the sample size (i.e., N is much larger
than n) the hypergeometric distribution is approximated reasonably well by a binomial
distribution with parameters n (number of trials) and p = D / N (probability of success in
a single trial).
HYPERGEOMETRIC FORMULA
When there are two groups of items such that there are ‘a’ items in the first group and
‘b’ items in the second group, so that the total number of items is (a + b), the probability
of selecting x items from the first group and (n-x) items from the second group is
C
C
a x . b nx
, where n is the total of items selected without replacement.
C
ab n
Examples:
1. A bag contains 3 blue chips and 3 green chips. If two chips are selected at
random, find the probability that both are blue.
Solution:
C
C
a x . b nx
; a = 3, b= 3, x=2, n=2, n-x=2-2=0
C
ab n
C
C
3 2 . 3 22 3 x1 1
0.2
The probability of both blue =
C
15
5
33 2
From the formula
2. A committee of 3 people is selected at random without replacement from a group
of 6 men and 3 women. Find the probability that the committee consists of 2 men
and 2 women.
Solution:
So into
a=6 b=3
n = 6+3=9
since the committee consists 2 men and 2 women
x=2
n-.x= 3-2=1
Pr 6C 2 3C1 15 x3 15
0.536
9C3
84
28
69
3. A group of 10 tanks contains 3 defective tanks. If 4 tanks are randomly selected
and tested find the probability that exactly one will be defective solution.
3 are defective
a=3
7 are good
b=7
Pr (one to be defective)
n=4
x=1
n-x=4-1=3
Pr (one to be exactly defective)
3C1 .7C3 105
0.5
10C 4
210
DO THIS
1. In a box of 10 shirts there are five (5) defective ones. If 5 shirts are sold at
random find the probability that exactly two are defective.
Answer
2. In a shipment of 12 lawn chairs 86 are brown and 4 are blue. If 3 chairs are sold
at random find the probability that that all are brown.
Answer Key:
1).
0.397
2)
0.255
Group Work
1. Revise the following probability questions and
answers
2. Discuss any problems encountered in the
computations of the probabilities.
1)
Find the probability of choosing 5 women from a committee of 15 women
P(choosing 5) =
1
1
=
C 5 3003
15
70
2)
What is the probability of drawing an ace or a spade from deck of playing
cards?
P( Ace)
4
P AUB P A P B P AUB
52
P spade
13
52
=
=
3)
4 13 1
52 52 52
16 4
52 13
There are problems pregnant for women. The probability of dying is
1
what is the
5
probability that at least one will die in every 5 women
1
P A
51
P A1
5
50
P(At least one will die) = use calculator
51
1 50
51 51
APPLICATION AND EXAMPLE
The classical application of the hypergeometric distribution is sampling without
replacement. Think of an urn with two types of marbles, black ones and white ones.
Define drawing a white marble as a success and drawing a black marble as a failure
(analogous to the binomial distribution). If the variable N describes the number of all
marbles in the urn (see contingency table above) and D describes the number of
white marbles (called defective in the example above), then N − D corresponds to the
number of black marbles.
Now, assume that there are 5 white and 45 black marbles in the urn. Standing next to
the urn, you close your eyes and draw 10 marbles without replacement. What's the
probability p (k=4) that you draw exactly 4 white marbles (and - of course - 6 black
marbles) ?
This problem is summarized by the following contingency table:
71
drawn
not drawn
total
white
marbles
4 (k)
1 = 5 − 4 (D − k)
5 (D)
black
marbles
6 = 10 − 4 (n − k)
39 = 50 + 4 − 10 − 5 (N + k − n − D)
45 (N − D)
total
10 (n)
40 (N − n)
50 (N)
The probability Pr (k = x) of drawing exactly x white marbles (= number of successes)
can be calculated by the formula
Hence, in this example x = 4, calculate
So, the probability of drawing exactly 4 white marbles is quite low (approximately 0.004)
and the event is very unlikely. It means, if you repeated your random experiment
(drawing 10 marbles from the urn of 50 marbles without replacement) 1000 times you
just would expect to obtain such a result 4 times.
But what about the probability of drawing even (all) 5 white marbles? You will intuitively
agree upon that this is even more unlikely than drawing 4 white marbles. Let us
calculate the probability for such an extreme event.
The contingency table is as follows:
drawn
white marbles 5 (k)
not drawn
0 = 5 − 5 (D − k)
total
5 (D)
black marbles 5 = 10 − 5 (n − k) 40 = 50 + 5 − 10 − 5 (N + k − n − D) 45 (N − D)
total 10 (n)
40 (N − n)
50 (N)
72
And we can calculate the probability as follows (notice that the denominator always
stays the same):
As expected, the probability of drawing 5 white marbles is even much lower than
drawing 4 white marbles.
Conclusion:
Consequently, one could expand the initial question as follows: If you draw 10 marbles
from an urn (containing 5 white and 45 black marbles), what's the probability of drawing
at least 4 white marbles? Or, what's the probability of drawing 4 white marbles and more
extreme outcomes such as drawing 5)? This corresponds to calculating the cumulative
probability p(k>=4) and can be calculated by the cumulative distribution function
(cdf). Since the hypergeometric distribution is a discrete probability distribution the
cumulative probability can be calculated easily by adding all corresponding single
probability values.
In our example you just have to sum-up Pr (k = 4) and Pr (k = 5):
Pr (k ≥ 4) = 0.003964583 + 0.0001189375 = 0.004083520
READ :
1. An Introduction to Probability & Random Processes By
Kenneth B & Gian-Carlo R, pages 184-195
BIVARIATE FREQUENCY DISTRIBUTIONS.
The bivariate normal distribution is the statistical distribution with probability function
73
where
and
is the correlation of and (Kenney and Keeping 1951, pp. 92 and 202-205; Whittaker
and Robinson 1967, p. 32)
are commonly used in place of
and
.
The marginal probabilities are then
=
=
And
=
=
74
JOINT PROBABILITY TABLES
This table is a correctly formatted joint probability table.
Days Listed Until Sold
Under 30
31-90
Over 90
Under $50,000
0.06
0.05
0.01
0.13
$50,000-99,999
0.03
0.19
0.10
0.31
$100,000-150,000
0.03
0.35
0.13
0.50
Over $150,000
0.01
0.04
0.01
0.06
Totals
0.13
0.63
0.25
1.00
Initial Asking Price
Totals
MARGINAL PROBABILITIES.
Let be partitioned into
disjoint sets
. Then the marginal probability of
and
is
where the general subset is denoted
READ :
1. An Introduction to Probability & Random Processes By
Kenneth B & Gian-Carlo R, pages 142-150
2. Exercise pg 150 Nos. 1,23,4,5,6,7,8,9,14,15,16,17,26.
75
REFLECTION: ICT resources are difficult to access!! The link opens up
avenue for Mathematics teachers to access ICT resources.
http://www.tsm-resources.com/suppl.html
76
UNIT 2 ( 40 HOURS):
RANDOM VARIABLES AND TEST DISTRIBUTIONS
MOMENTS
The probability distribution of a random variable is often characterised by a small
number of parameters, which also have a practical interpretation. For example, it is
often enough to know what its "average value" is. This is captured by the mathematical
concept of expected value of a random variable, denoted E[X]. Note that in general,
E[f(X)] is not the same as f(E[X]). Once the "average value" is known, one could then
ask how far from this average value the values of X typically are, a question that is
answered by the variance and standard deviation of a random variable.
Mathematically, this is known as the (generalised) problem of moments: for a given
class of random variables X, find a collection {fi} of functions such that the expectation
values E[fi(X)] fully characterize the distribution of the random variable X.
Equivalence of random variables
There are several different senses in which random variables can be considered to be
equivalent. Two random variables can be equal, equal almost surely, equal in mean, or
equal in distribution.
In increasing order of strength, the precise definition of these notions of equivalence is
given below.
Equality in distribution
Two random variables X and Y are equal in distribution if they have the same
distribution functions:
Two random variables having equal moment generating functions have the same
distribution.
Equality in mean
Two random variables X and Y are equal in p-th mean if the pth moment of |X − Y| is
zero, that is,
Equality in pth mean implies equality in qth mean for all q<p. As in the previous case,
there is a related distance between the random variables, namely
77
Equality
Finally, the two random variables X and Y are equal if they are equal as functions on
their probability space, that is,
MOMENT-GENERATING FUNCTION
In probability theory and statistics, the moment-generating function of a random
variable X is
wherever this expectation exists. The moment-generating function generates the
moments of the probability distribution.
For vector-valued random variables X with real components, the moment-generating
function is given by
where t is a vector and
is the dot product.
Provided the moment-generating function exists in an interval around t = 0, the nth
moment is given by
If X has a continuous probability density function f(x) then the moment generating
function is given by
where mi is the ith moment. MX( − t) is just the two-sided Laplace transform of f(x).
78
Regardless of whether the probability distribution is continuous or not, the momentgenerating function is given by the Riemann-Stieltjes integral
where F is the cumulative distribution function.
If X1, X2, ..., Xn is a sequence of independent (and not necessarily identically distributed)
random variables, and
where the ai are constants, then the probability density function for Sn is the convolution
of the probability density functions of each of the Xi and the moment-generating function
for Sn is given by
Related to the moment-generating function are a number of other transforms that are
common in probability theory, including the characteristic function and the probabilitygenerating function.
MARKOV'S INEQUALITY
f(x)
XX | f ( x)
Markov's inequality gives an upper bound for the probability that X lies within
XX | f ( x)
In probability theory, Markov's inequality gives an upper bound for the probability that
a non-negative function of a random variable is greater than or equal to some positive
constant. It is named after the Russian mathematician Andrey Markov, although it
appeared earlier in the work of Pafnuty Chebyshev (Markov's teacher).
79
Markov's inequality (and other similar inequalities) relate probabilities to expectations,
and provide (frequently) loose but still useful bounds for the cumulative distribution
function of a random variable.
Special case: probability theory
For any event E, let IE be the indicator random variable of E, that is, IE = 1 if E occurs
and = 0 otherwise. Thus I(|X| ≥ a) = 1 if the event |X| ≥ a occurs, and I(|X| ≥ a) = 0 if |X| < a.
Then, given a>0,
Therefore
Now observe that the left side of this inequality is the same as
Thus we have
and since a > 0, we can divide both sides by a.
READ :
1. Robert B. Ash, Lectures on Statistics, page 9-13
2. An Introduction to Probability & Random Processes By Kenneth
B & Gian-Carlo R, pages 366 -374 & 404 - 407
Exercise on pg 376 -376 Nos. 1,3,7,8
Exercise on pg 442 Nos. 1,2,3,4,5
Ref:
http://en.wikipedia.org/wiki/Moment-generating_function
http://en.wikipedia.org/wiki/characteristic_function_%28probabil
ity_theory%29.
http://en.wikipedia.org/wiki/Integral_transform
80
CHEBYSHEV'S INEQUALITY
In probability theory, Chebyshev's inequality (also known as Tchebysheff's
inequality, Chebyshev's theorem, or the Bienaymé-Chebyshev inequality), named
after Pafnuty Chebyshev, who first proved it, states that in any data sample or
probability distribution, nearly all the values are close to the mean value, and provides a
quantitative description of "nearly all" and "close to". For example, no more than 1/4 of
the values are more than 2 standard deviations away from the mean, no more than 1/9
are more than 3 standard deviations away, no more than 1/25 are more than 5 standard
deviations away, and so on.
Probabilistic statement
Let X be a random variable with expected value μ and finite variance σ2. Then for any
real number k > 0,
Only the cases k > 1 provide useful information.
As an example, using k=√2 shows that at least half of the values lie in the interval (μ −
√2 σ, μ + √2 σ).
Typically, the theorem will provide rather loose bounds. However, the bounds provided
by Chebyshev's inequality cannot, in general (remaining sound for variables of arbitrary
distribution), be improved upon. For example, for any k > 1, the following example
(where σ = 1/k) meets the bounds exactly.
The theorem can be useful despite loose bounds because it applies to random variables
of any distribution, and because these bounds can be calculated knowing no more
about the distribution than the mean and variance.
Chebyshev's inequality is used for proving the weak law of large numbers.
81
Example application
For illustration, assume we have a large body of text, for example articles from a
publication. Assume we know that the articles are on average 1000 characters long with
a standard deviation of 200 characters. From Chebyshev's inequality we can then
deduce that at least 75% of the articles have a length between 600 and 1400 characters
(k = 2).
Probabilistic proof
Markov's inequality states that for any real-valued random variable Y and any positive
number a, we have Pr(|Y| > a) ≤ E(|Y|)/a. One way to prove Chebyshev's inequality is to
apply Markov's inequality to the random variable Y = (X − μ)2 with a = (σk)2.
It can also be proved directly. For any event A, let IA be the indicator random variable of
A, i.e. IA equals 1 if A occurs and 0 otherwise. Then
The direct proof shows why the bounds are quite loose in typical cases: the number 1 to
the left of "≥" is replaced by [(X − μ)/(kσ)]2 to the right of "≥" whenever the latter exceeds
1. In some cases it exceeds 1 by a very wide margin.
READ :
1. An Introduction to Probability & Random Processes By Kenneth
B & Gian-Carlo R, pages 305-318
Exercise on pg 309 nos. 1,2,3,4,5
Exercise on pg 320-324 Nos. 1,3,10,12
CORRELATION TYPES
Correlation is a measure of association between two variables. The variables are not
designated as dependent or independent. The two most popular correlation coefficients
are: Spearman's correlation coefficient rho and Pearson's product-moment correlation
coefficient.
When calculating a correlation coefficient for ordinal data, select Spearman's technique.
For interval or ratio-type data, use Pearson's technique.
82
The value of a correlation coefficient can vary from minus one to plus one. A minus one
indicates a perfect negative correlation, while a plus one indicates a perfect positive
correlation. A correlation of zero means there is no relationship between the two
variables. When there is a negative correlation between two variables, as the value of
one variable increases, the value of the other variable decreases, and vise versa. In
other words, for a negative correlation, the variables work opposite each other. When
there is a positive correlation between two variables, as the value of one variable
increases, the value of the other variable also increases. The variables move together.
The standard error of a correlation coefficient is used to determine the confidence
intervals around a true correlation of zero. If your correlation coefficient falls outside of
this range, then it is significantly different than zero. The standard error can be
calculated for interval or ratio-type data (i.e., only for Pearson's product-moment
correlation).
The significance (probability) of the correlation coefficient is determined from the tstatistic. The probability of the t-statistic indicates whether the observed correlation
coefficient occurred by chance if the true correlation is zero. In other words, it asks if the
correlation is significantly different than zero. When the t-statistic is calculated for
Spearman's rank-difference correlation coefficient, there must be at least 30 cases
before the t-distribution can be used to determine the probability. If there are fewer than
30 cases, you must refer to a special table to find the probability of the correlation
coefficient.
Example
A company wanted to know if there is a significant relationship between the total
number of salespeople and the total number of sales. They collect data for five months.
Variable 1 Variable 2
207
6907
180
5991
220
6810
205
6553
190
6190
Correlation coefficient = .921
Standard error of the coefficient = ..068
t-test for the significance of the coefficient = 4.100
Degrees of freedom = 3
Two-tailed probability = .0263
83
Another Example
Respondents to a survey were asked to judge the quality of a product on a four-point
Likert scale (excellent, good, fair, poor). They were also asked to judge the reputation of
the company that made the product on a three-point scale (good, fair, poor). Is there a
significant relationship between respondents perceptions of the company and their
perceptions of quality of the product?
Since both variables are ordinal, Spearman's method is chosen. The first variable is the
rating for the quality the product. Responses are coded as 4=excellent, 3=good, 2=fair,
and 1=poor. The second variable is the perceived reputation of the company and is
coded 3=good, 2=fair, and 1=poor.
Variable 1
Variable 2
4
3
2
2
1
2
3
3
4
3
1
1
2
1
Correlation coefficient rho = .830
t-test for the significance of the coefficient = 3.332
Number of data pairs = 7
Probability must be determined from a table because of the small sample size.
REGRESSION
Simple regression is used to examine the relationship between one dependent and one
independent variable. After performing an analysis, the regression statistics can be
used to predict the dependent variable when the independent variable is known.
Regression goes beyond correlation by adding prediction capabilities.
People use regression on an intuitive level every day. In business, a well-dressed man
is thought to be financially successful. A mother knows that more sugar in her children's
diet results in higher energy levels. The ease of waking up in the morning often depends
84
on how late you went to bed the night before. Quantitative regression adds precision by
developing a mathematical formula that can be used for predictive purposes.
For example, a medical researcher might want to use body weight (independent
variable) to predict the most appropriate dose for a new drug (dependent variable). The
purpose of running the regression is to find a formula that fits the relationship between
the two variables. Then you can use that formula to predict values for the dependent
variable when only the independent variable is known. A doctor could prescribe the
proper dose based on a person's body weight.
The regression line (known as the least squares line) is a plot of the expected value of
the dependent variable for all values of the independent variable. Technically, it is the
line that "minimizes the squared residuals". The regression line is the one that best fits
the data on a scatterplot.
Using the regression equation, the dependent variable may be predicted from the
independent variable. The slope of the regression line (b) is defined as the rise divided
by the run. The y intercept (a) is the point on the y axis where the regression line would
intercept the y axis. The slope and y intercept are incorporated into the regression
equation. The intercept is usually called the constant, and the slope is referred to as the
coefficient. Since the regression model is usually not a perfect predictor, there is also an
error term in the equation.
In the regression equation, y is always the dependent variable and x is always the
independent variable. Here are three equivalent ways to mathematically describe a
linear regression model.
y = intercept + (slope x) + error
y = constant + (coefficient x) + error
y = a + bx + e
The significance of the slope of the regression line is determined from the t-statistic. It is
the probability that the observed correlation coefficient occurred by chance if the true
correlation is zero. Some researchers prefer to report the F-ratio instead of the tstatistic. The F-ratio is equal to the t-statistic squared.
The t-statistic for the significance of the slope is essentially a test to determine if the
regression model (equation) is usable. If the slope is significantly different than zero,
then we can use the regression model to predict the dependent variable for any value of
the independent variable.
On the other hand, take an example where the slope is zero. It has no prediction ability
because for every value of the independent variable, the prediction for the dependent
variable would be the same. Knowing the value of the independent variable would not
improve our ability to predict the dependent variable. Thus, if the slope is not
significantly different than zero, don't use the model to make predictions.
85
The coefficient of determination (r-squared) is the square of the correlation coefficient.
Its value may vary from zero to one. It has the advantage over the correlation coefficient
in that it may be interpreted directly as the proportion of variance in the dependent
variable that can be accounted for by the regression equation. For example, an rsquared value of .49 means that 49% of the variance in the dependent variable can be
explained by the regression equation. The other 51% is unexplained.
The standard error of the estimate for regression measures the amount of variability in
the points around the regression line. It is the standard deviation of the data points as
they are distributed around the regression line. The standard error of the estimate can
be used to develop confidence intervals around a prediction.
Example
A company wants to know if there is a significant relationship between its advertising
expenditures and its sales volume. The independent variable is advertising budget and
the dependent variable is sales volume. A lag time of one month will be used because
sales are expected to lag behind actual advertising expenditures. Data was collected for
a six month period. All figures are in thousands of dollars. Is there a significant
relationship between advertising budget and sales volume?
Independent Variable
Dependent Variable
4.2
27.1
6.1
30.4
3.9
25.0
5.7
29.7
7.3
40.1
5.9
28.8
Model: y = 10.079 + (3.700 x) + error
Standard error of the estimate = 2.568
t-test for the significance of the slope = 4.095
Degrees of freedom = 4
Two-tailed probability = .0149
r-squared = .807
You might make a statement in a report like this: A simple linear regression was
performed on six months of data to determine if there was a significant relationship
between advertising expenditures and sales volume. The t-statistic for the slope was
significant at the .05 critical alpha level, t(4)=4.10, p=.015. Thus, we reject the null
86
hypothesis and conclude that there was a positive significant relationship between
advertising expenditures and sales volume. Furthermore, 80.7% of the variability in
sales volume could be explained
READ :
1) An Introduction to Probability & Random Processes By Kenneth
B & Gian-Carlo R, pages 18-30, 212-215, 300-303
2) Robert B. Ash, Lectures on Statistics, page 28-29.
Ref: "http://en.wikipedia.org/wiki/Correlation
Ref: "http://en.wikipedia.org/wiki/Regression
CHI-SQUARE TEST
A chi-square test is any statistical hypothesis test in which the test statistic has a chisquare distribution when the null hypothesis is true, or any in which the probability
distribution of the test statistic (assuming the null hypothesis is true) can be made to
approximate a chi-square distribution as closely as desired by making the sample size
large enough.
Specifically, a chi-square test for independence evaluates statistically significant
differences between proportions for two or more groups in a data set.
Pearson's chi-square test, also known as the Chi-square goodness-of-fit test
Yates' chi-square test also known as Yates' correction for continuity
Mantel-Haenszel chi-square test
Linear-by-linear association chi-square test
In probability theory and statistics, the chi-square distribution (also chi-squared or χ2
distribution) is one of the most widely used theoretical probability distributions in
inferential statistics, i.e. in statistical significance tests. It is useful because, under
reasonable assumptions, easily calculated quantities can be proven to have
distributions that approximate to the chi-square distribution if the null hypothesis is true.
87
If Xi are k independent, normally distributed random variables with mean 0 and variance
1, then the random variable
is distributed according to the chi-square distribution. This is usually written
The chi-square distribution has one parameter: k - a positive integer that specifies the
number of degrees of freedom (i.e. the number of Xi)
The chi-square distribution is a special case of the gamma distribution.
The best-known situations in which the chi-square distribution is used are the common
chi-square tests for goodness of fit of an observed distribution to a theoretical one, and
of the independence of two criteria of classification of qualitative data. However, many
other statistical tests lead to a use of this distribution.
CHARACTERISTIC FUNCTION
The characteristic function of the Chi-square distribution is
Properties
The chi-square distribution has numerous applications in inferential statistics, for
instance in chi-square tests and in estimating variances. It enters the problem of
estimating the mean of a normally distributed population and the problem of estimating
the slope of a regression line via its role in Student's t-distribution. It enters all analysis
of variance problems via its role in the F-distribution, which is the distribution of the ratio
of two independent chi-squared random variables divided by their respective degrees of
freedom.
88
Various chi and chi-square distributions
Name
Statistic
chi-square distribution
noncentral chi-square distribution
chi distribution
noncentral chi distribution
READ :
Ref: http://en.wikipedia.org/wiki/pearson%chi-square_test
Ref: http://en.wikipedia.org/wiki/Chi-Square_test
STUDENT'S T-TEST
A t test is any statistical hypothesis test for two groups in which the test statistic has a
Student's t distribution if the null hypothesis is true.
History
The t statistic was introduced by William Sealy Gosset for cheaply monitoring the quality
of beer brews. "Student" was his pen name. Gosset was a statistician for the Guinness
brewery in Dublin, Ireland, and was hired due to Claude Guinness's innovative policy of
recruiting the best graduates from Oxford and Cambridge to apply biochemistry and
statistics to Guinness' industrial processes. Gosset published the t test in Biometrika in
1908, but was forced to use a pen name by his employer who regarded the fact that
they were using statistics as a trade secret. In fact, Gosset's identity was unknown not
only to fellow statisticians but to his employer—the company insisted on the pseudonym
so that it could turn a blind eye to the breach of its rules.
Today, it is more generally applied to the confidence that can be placed in judgments
made from small samples.
89
Use
Among the most frequently used t tests are:
A test of the null hypothesis that the means of two normally distributed
populations are equal. Given two data sets, each characterized by its mean,
standard deviation and number of data points, we can use some kind of t test to
determine whether the means are distinct, provided that the underlying
distributions can be assumed to be normal. All such tests are usually called
Student's t tests, though strictly speaking that name should only be used if the
variances of the two populations are also assumed to be equal; the form of the
test used when this assumption is dropped is sometimes called Welch's t test.
There are different versions of the t test depending on whether the two samples
are
o independent of each other (e.g., individuals randomly assigned into two
groups), or
o paired, so that each member of one sample has a unique relationship with
a particular member of the other sample (e.g., the same people measured
before and after an intervention, or IQ test scores of a husband and wife).
If the t value that is calculated is above the threshold chosen for statistical
significance (usually the 0.05 level), then the null hypothesis that the two groups
do not differ is rejected in favor of an alternative hypothesis, which typically
states that the groups do differ.
A test of whether the mean of a normally distributed population has a value
specified in a null hypothesis.
A test of whether the slope of a regression line differs significantly from 0.
Once a t value is determined, a P value can be found using a table of values from
Student's t-distribution.
Confidence intervals using a small sample size
Consider a normally distributed population. To estimate the population's variance take a
sample of size n and calculate the sample's variance, s. An unbiased estimator of the
population's variance is
Clearly for small values of n this estimation is inaccurate. Hence for samples of small
size instead of calculating the z value for the number of standard deviations from the
mean
90
and using probabilities based on the normal distribution, calculate the t value
The probability that the t value is within a particular interval may be found using the t
distribution. The sample's degrees of freedom are the number of data that need to be
known before the rest of the data can be calculated.
e.g.
A random sample of things have weights
30.02, 29.99, 30.11, 29.97, 30.01, 29.99
Calculate a 95% confidence interval for the population's mean weight.
Assume the population ~ N(μ,σ2)
The samples' mean weight is 30.015 with standard deviation of 0.045. With the mean
and the first five weights it is possible to calculate the sixth weight. Consequently there
are five degrees of freedom.
The t distribution tells us that, for five degrees of freedom, the probability that t > 2.571
is 0.025. Also, the probability that t < −2.571 is 0.025. Using the formula for t with t = ±
2.571 a 95% confidence interval for the populations mean may be found by making μ
the subject of the equation.
i.e.
(29.97 < μ < 30.06)
READ :
1. Introduction to Probability By Charles M. Grinstead, pages 1830, 212-215, 300-303
2. Robert B. Ash, Lectures on Statistics, page 23-29.
Answer problems 1- 6 on pg 23.
Ref: http://en.wikipedia.org/wiki/Statistical_Hypothesis_testing
Ref: http://en.wikipedia.org/wiki/Null_hypothesis
91
REFLECTION: The study of Correlation, Regression Hypothesis testing
and other Mathematical modelling maybe simplified through ICT. The
following link enables trainees to learn modelling with ease
http://www.ncaction.org.uk/subjects/maths/ict-lrn.htm
92
UNIT 3 ( 40 HOURS): PROBABILITY THEORY
INDICATOR FUNCTION
In mathematics, an indicator function or a characteristic function is a function
defined on a set X that indicates membership of an element in a subset A of X.
The indicator function of a subset A of a set X is a function
defined as
The indicator function of A is sometimes denoted
χA(x) or
or even A(x).
BONFERONI INEQUALITY
Let
be the probability that is true, and
be the probability that at least one
of , , ..., is true. Then "the" Bonferroni inequality, also known as Boole's
inequality, states that
where denotes the union. If and are disjoint sets for all and , then the inequality
becomes an equality. A beautiful theorem that expresses the exact relationship between
the probability of unions and probabilities of individual events is known as the inclusionexclusion principle.
A slightly wider class of inequalities are also known as "Bonferroni inequalities."
93
GENERATING FUNCTION
In mathematics a generating function is a formal power series whose coefficients
encode information about a sequence an that is indexed by the natural numbers.
There are various types of generating functions, including ordinary generating
functions, exponential generating functions, Lambert series, Bell series, and
Dirichlet series; definitions and examples are given below. Every sequence has a
generating function of each type. The particular generating function that is most useful
in a given context will depend upon the nature of the sequence and the details of the
problem being addressed.
Generating functions are often expressed in closed form as functions of a formal
argument x. Sometimes a generating function is evaluated at a specific value of x.
However, it must be remembered that generating functions are formal power series, and
they will not necessarily converge for all values of x.
If an is the probability mass function of a discrete random variable, then its ordinary
generating function is called a probability-generating function.
The ordinary generating function can be generalised to sequences with multiple
indexes. For example, the ordinary generating function of a sequence am,n (where n and
m are natural numbers) is
CHARACTERISTIC FUNCTION (PROBABILITY THEORY)
In probability theory, the characteristic function of any random variable completely
defines its probability distribution. On the real line it is given by the following formula,
where X is any random variable with the distribution in question:
where t is a real number, i is the imaginary unit, and E denotes the expected value.
If FX is the cumulative distribution function, then the characteristic function is given by
the Riemann-Stieltjes integral
94
In cases in which there is a probability density function, fX, this becomes
If X is a vector-valued random variable, one takes the argument t to be a vector and tX
to be a dot product.
Every probability distribution on R or on Rn has a characteristic function, because one is
integrating a bounded function over a space whose measure is finite.
The continuity theorem
If the sequence of characteristic functions of distributions Fn converges to the
characteristic function of a distribution F, then Fn(x) converges to F(x) at every value of x
at which F is continuous.
USES OF CHARACTERISTIC FUNCTIONS
Characteristic functions are particularly useful for dealing with functions of independent
random variables. For example, if X1, X2, ..., Xn is a sequence of independent (and not
necessarily identically distributed) random variables, and
where the ai are constants, then the characteristic function for Sn is given by
In particular,
characteristic function:
. To see this, write out the definition of
.
Observe that the independence of X and Y is required to establish the equality of the
third and fourth expressions.
95
Because of the continuity theorem, characteristic functions are used in the most
frequently seen proof of the central limit theorem.
Characteristic functions can also be used to find moments of random variable. Provided
that nth moment exists, characteristic function can be differentiated n times and
READ
1. Robert B. Ash, Lectures on Statistics, page 32 of 45:
Ref
"http://en.wikipedia.org/wiki/Characteristic_function_%28probability_
theory%29
STATISTICAL INDEPENDENCE
In probability theory, to say that two events are independent intuitively means that the
occurrence of one event makes it neither more nor less probable that the other occurs.
For example:
The event of getting a "6" the first time a die is rolled and the event of getting a
"6" the second time are independent.
By contrast, the event of getting a "6" the first time a die is rolled and the event
that the sum of the numbers seen on the first and second trials is "8" are
dependent.
If two cards are drawn with replacement from a deck of cards, the event of
drawing a red card on the first trial and that of drawing a red card on the second
trial are independent.
By contrast, if two cards are drawn without replacement from a deck of cards,
the event of drawing a red card on the first trial and that of drawing a red card on
the second trial are dependent.
Similarly, two random variables are independent if the conditional probability distribution
of either given the observed value of the other is the same as if the other's value had
not been observed.
96
INDEPENDENT EVENTS
The standard definition says:
Two events A and B are independent if and only if Pr(A ∩ B) = Pr(A)Pr(B).
Here A ∩ B is the intersection of A and B, that is, it is the event that both events A and B
occur.
More generally, any collection of events -- possibly more than just two of them -- are
mutually independent if and only if for any finite subset A1, ..., An of the collection we
have
This is called the multiplication rule for independent events.
If two events A and B are independent, then the conditional probability of A given B is
the same as the "unconditional" (or "marginal") probability of A, that is,
There are at least two reasons why this statement is not taken to be the definition of
independence: (1) the two events A and B do not play symmetrical roles in this
statement, and (2) problems arise with this statement when events of probability 0 are
involved.
When one recalls that the conditional probability Pr(A | B) is given by
(so long as Pr(B) ≠ 0 )
one sees that the statement above is equivalent to
which is the standard definition given above.
RANDOM SAMPLE
A sample is a subset chosen from a population for investigation. A random sample is
one chosen by a method involving an unpredictable component. Random sampling can
also refer to taking a number of independent observations from the same probability
97
distribution, without involving any real population. A probability sample is one in which
each item has a known probability of being in the sample.
The sample will usually not be completely representative of the population from which it
was drawn— this random variation in the results is known as sampling error. In the case
of random samples, mathematical theory is available to assess the sampling error.
Thus, estimates obtained from random samples can be accompanied by measures of
the uncertainty associated with the estimate. This can take the form of a standard error,
or if the sample is large enough for the central limit theorem to take effect, confience
intervals may be calculated.
Types of random sample
A simple random sample is selected so that every possible sample has an equal
chance of being selected.
A self-weighting sample, also known as an epsem sample, is one in which every
individual, or object, in the population of interest has an equal opportunity of
being selected for the sample. Simple random samples are self-weighting.
Stratified sampling involves selecting independent samples from a number of
subpopulations (or strata) within the population. Great gains in efficiency are
sometimes possible from judicious stratification.
Cluster sampling involves selecting the sample units in groups. For example, a
sample of telephone calls may be collected by first taking a collection of
telephone lines and collecting all the calls on the sampled lines. The analysis of
cluster samples must take into account the intra-cluster correlation which reflects
the fact that units in the same cluster are likely to be more similar than two units
picked at random.
MULTINOMIAL DISTRIBUTION
In probability theory, the multinomial distribution is a generalization of the binomial
distribution.
The binomial distribution is the probability distribution of the number of "successes" in n
independent Bernoulli trials, with the same probability of "success" on each trial. In a
multinomial distribution, each trial results in exactly one of some fixed finite number k of
possible outcomes, with probabilities p1, ..., pk (so that pi ≥ 0 for i = 1, ..., k and
), and there are n independent trials. Then let the random variables Xi
indicate the number of times outcome number i was observed over the n trials.
follows a multinomial distribution with parameters n and p.
98
SOLUTIONS FROM MULTINOMIAL DISTRIBUTION FORMULA
A short version of the multinomial formula for three consecutive outcomes is given
below.
If X consists of events E1, E2, E3, which have the corresponding probabilities of p 1, p2,
and p3 of occurring, where x1 is the number of times E1 will occur, x2 is the number of
times E2 will occur, and x3 is the number of times E3 will occur, then the probability of X
is
n!
.
x1 ! x 2 ! x3 !
p
x1
1
x2
x3
2
3
.p .p
where x1 + x2 + x3 = n and p1 + p2 + p3 = 1
Example:
1) In a large city, 60% of the workers drive to work, 30% take the bus, and 10% take
the train. If 5 workers are selected at random, find the probability that 2 will drive, 2
will take the us, and 1 will take the train.
Solution:
n= 5, x1=2, x2 = 2, x3= 1 and p1=0.6, p2= 0.3, and p3 = 0.1
Hence, the probability that 2 workers will take the bus, and one will take the train is
2
2
1
5!
. (0.6) (0.3) (0.1) 0.0972
2 ! 2 !1 !
2) A box contains 5 red balls, 3 blue balls, and 2 white balls. If 4 balls are selected with
replacement, find the probability of getting 2 red balls, one blue ball, and one white
ball.
Solution:
5
3
2
n=4, x1=2, x2=1, x3=1, and p1= , p2=
, and p3=
. Hence, the probability of
10
10
10
getting 2 red balls, one blue ball, and one white ball is
2
4! 5 3
2! 1! 1! 10 10
{ Allan G, 2005,
1
1
2
3 9
0.18
12
10
200 50
pg 132}
ORDER STATISTIC
Probability distributions for the n = 5 order statistics of an exponential distribution with θ
= 3.
In statistics, the kth order statistic of a statistical sample is equal its kth-smallest value.
Together with rank statistics, order statistics are among the most fundamental tools in
non-parametric statistics and inference.
99
Important special cases of the order statistics are the minimum and maximum value of a
sample, and (with some qualifications discussed below) the sample median and other
sample quartiles.
When using probability theory to analyse order statistics of random samples from a
continuous distribution, the cumulative distribution function is used to reduce the
analysis to the case of order statistics of the uniform distribution.
READ :
1. Robert B. Ash, Lectures on Statistics, page 25 -26 and Answer
problems 1-4 on pg 26/27.
Ref: http://en.wikipedia.org/wiki/probability _distribution
Ref: http://en.wikipedia.org/wiki/Ranking
Ref: http://en.wikipedia.org/wiki/non-parametric_Statistics
Notation and examples
For example, suppose that four numbers are observed or recorded, resulting in a
sample of size n = 4. If the sample values are
6, 9, 3, 8,
they will usually be denoted
where the subscript i in xi indicates simply the order in which the observations were
recorded and is usually assumed not to be significant. A case when the order is
significant is when the observations are part of a time series.
The order statistics would be denoted
where the subscript (i) enclosed in parentheses indicates the ith order statistic of the
sample.
The first order statistic (or smallest order statistic) is always the minimum of the
sample, that is,
100
where, following a common convention, we use upper-case letters to refer to random
variables, and lower-case letters (as above) to refer to their actual observed values.
Similarly, for a sample of size n, the nth order statistic (or largest order statistic) is the
maximum, that is,
The sample range is the difference between the maximum and minimum. It is clearly a
function of the order statistics:
A similar important statistic in exploratory data analysis that is simply related to the
order statistics is the sample interquartile range.
The sample median may or may not be an order statistic, since there is a single middle
value only when the number n of observations is odd. More precisely, if n = 2m + 1 for
some m, then the sample median is X(m + 1) and so is an order statistic. On the other
hand, when n is even, n = 2m and there are two middle values, X(m) and X(m + 1), and the
sample median is some function of the two (usually the average) and hence not an
order statistic. Similar remarks apply to all sample quantiles.
MULTIVARIATE NORMAL DISTRIBUTION
In probability theory and statistics, a multivariate normal distribution, also sometimes
called a multivariate Gaussian distribution, is a specific probability distribution, which
can be thought of as a generalization to higher dimensions of the one-dimensional
normal distribution (also called a Gaussian distribution).
Higher moments
The kth-order moments of X are defined by
where
The central k-order moments are given as follows
(a) If k is odd,
.
(b) If k is even with k = 2λ, then
101
where the sum is taken over all allocations of the set
into λ (unordered)
pairs, giving (2λ − 1)! / (2λ − 1(λ − 1)!) terms in the sum, each being the product of λ
covariances. The covariances are determined by replacing the terms of the list
by the corresponding terms of the list consisting of r1 ones, then r2 twos,
etc, after each of the possible allocations of the former list into pairs.
In particular, the 4-order moments are
For fourth order moments (four variables) there are three terms. For sixth-order
moments there are 3 × 5 = 15 terms, and for eighth-order moments there are 3 × 5 × 7
= 105 terms.
102
SYNTHESIS OF THE MODULE
At the end of this module learners are expected to compute various measures of
dispersions and apply the laws of probability to various probability distributions. The
learners should be able to solve various coefficients of correlation and regression. Unit
one of Probability and Statistics covers Frequency distributions relative and cumulative
distributions, various frequency curves, mean, Mode Median. Quartiles and Percentiles,
Standard deviation, symmetrical and skewed distributions. The learner is introduced to
various statistical measures and guided examples.
The examples are well illustrated and learners can follow without difficulty. It is
recommended that learners attempt the formative evaluations given to assess their
progress in the understanding of the content. Learners should take time to check the
recommended reference material in CD’S, attached open source materials and the
recommended websites. Most importantly, learners are encouraged to read the content
widely and attempt the questions after each topic. Unit two of the module takes learners
through Moment and moment generating function, Markov and Chebychev inequalities,
special Univariate distributions, Bivariate probability distribution; Joint Marginal and
conditional distributions; Independence; Bivariate expectation Regression and
Correlation; Calculation of regression and correlation coefficient for bivariate data.
Distribution function of random variables, Bivariate normal distribution. Derived
distributions such as Chi-Square, t and F.
Unit two has various learning activities to aid learning and learners are advised to
master the content of the various sub-topics and assess themselves through the
formative evaluations. Failure to answer the formative assessments should be a positive
indicator that learners should revise the sub-topics before progressing to other subtopics. The tasks given under the different learning activities demands that you
demonstrate a high level of ICT skills competency. The learning objectives are well
stated in the beginning of the module and should guide learners in the level of
expectations of the module.
Unit three focuses on probability theory and concentrates on the various probability
distributions.
The summative evaluation will be used to judge the learners mastery of the module. It is
recommended that learners revise the module before sitting for the final summative
evaluation.
103
SUMMATIVE EVALUATION
Answer Any Four Questions. Each question carries 15 marks.
Question 1: General Statistics
1) In the following table, the weights of 40 cows are recorded to the nearest kilogram.
128 161 135 142 145 156 150 145
157 138 150 147 140 125 144 173
144 146 140 176 154 148 163 164
135 146 142 142 149 119 134 158
165 168 138 147 152 153 136 126
Find;
a). the highest weight
b). the least weight
c). the range
d). construct a frequency distribution table starting with a class of 118-126
e). calculate the mean of the data
f). calculate the standard deviation
Question 2: General Probability
2) A). A coin and a die are thrown together. Draw a possibility space diagram and find
the probability of obtaining:
a). a head
b). a number greater than 4
c). a head and a number greater than 4
d). a head or a number greater than 4
19
2
4
B). Events M and N are such that P(M) =
, P(N) = and P(M N)= . Find P(M
5
5
30
N)
.
Question 3: Poisson Distribution
3) A book contains 500 pages and has 750 misprints.
a). What is the average number of misprints per page?
b) Find the probability that page 427 contains
i).
no misprints
ii).
exactly 4 misprints
c). find the probability that pages 427 and 428 will contain no misprints
104
Question 4: Continuous Random Variable
4) A continuous random variable (r.v) X has a probability density function (p.d.f) f(x)
where
f ( x)
k ( x 2) 2 2 x 0
4k
0 x 1 13
0
otherwise
a) Find the value of the constant k
b) Sketch y=f(x)
c) Find P( - 1 X 1)
d) Find P(x>1)
Probability of an event
5). Given that P(AUB) =7/8, P(A B)=1/4 and P(A’)=5/8, find the values of
a)
P(A)
b)
P(B)
c)
P(A B’)
d)
P(A’U B’)
e)
The probability that only one of A, B will occur.
6).
Expected Value
The continuous random variable r.v has the p.d.f
1
f ( x) x
0 x 1
2
Find:
a).
b).
c).
7).
E(X)
E(24X +6)
E( 1-X)
1
2
The masses, to the nearest kg, of 50 boys are recorded below.
Mass (kg)
Frequency (f)
a).
b).
60-64 65-69 70-74 75-79 80-84 85-89
2
6
12
14
10
6
Construct a cumulative frequency curve
Use the curve to estimate the ;
i)
Median
ii).
Interquartile range
iii).
7th decile
iii).
60th percentile.
105
MARKING SCHEME OF SUMMATIVE EVALUATION
1).
a)
176
b)
119
c)
176-119=57
d)
Using 7 classes gives us a class interval of 9
Weight(kg) Tally
Frequency
118-126
///
3
127-135
////
5
136-144
//// ////
9
145-153
//// //// // 12
154-162
////
5
163-171
////
4
172-180
//
2
Total 40
e) Accept any method of calculating the mean
f). Accept any method of calculating the standard deviation
2)
A). A coin has either Head(H) or Tail(T) while a die has sides 1,2,3,4,5&6.
Coin / Die 1
2
3
4
5
6
Coin H
H1 H2 H3 H4 H5 H6
Coin T
T1 T2 T3 T4 T5 T6
Sample space=12.
a).
6/12=1/2
b).
4/12=1/3
c).
2/12=1/6
d).
8/12=2/3
106
B) . P(M N)= P(M)+P(N)-P(M N).
4
19
2
=
+
- P(M N).
5
5
30
19 12
24
7
+
P(M N) =
30
30
30 30
3).
a) average number per page = 750/500=1.5
b) Let X be ‘the number of misprints per page’. Then, assuming that misprints
occur at random, X ~ P0(1.5)
i).
P(X= 0) = e-1.5
= 0.2231…
P(there will be no misprints on page 427) = 0.223 ( 3d.p).
(1.5) 4
ii).
P(X=4)= e-1.5
= 0.0470…
4!
P( there will be 4 misprints on page 427) = 0.047 ( 3d.p)
c). We expect 1.5 misprints on each page and so on two pages 427 & 428 we
expect 1.5 + 1.5 = 3 misprints.
Let Y be the ‘’number of misprints on two pages’’
Y ~ P(3), so P0(Y=0)= e-3
= 0.4421
4).
a). Since X is a random variable, then
f ( x)dx 1
all x
1
3
4kdx 1
0
1
0
2
k ( x 2) dx
2
11
0
k
3
( x 2)
4k x 03 1
3
2
k
4
(8) 4k 1
3
3
8k=1
1
k=
8
b) The p.d.f of X is
Therefore
f ( x)
1
( x 2) 2
8
1
2
0
2 x 0
0 x 1 13
otherwise
107
y
1
2
y
x
-2
0
11
3
c)
y
1
2
y
x
-2
-1
0
0
P(- 1 x 0 )
1
8 ( x 2)
11
2
1
dx
3
7
24
and
P(0 x 1) area of
rec tan gle
1
2
Therefore
P (1 X 1)
7 1 19
24 2 24
108
d).
y
1
2
y
x
-2
0
P(0 X 1) = area of rectangle=
Therefore P(x>1) =
5)
a)
b)
c)
d)
7).
1
3
1 1 1
.
3 2 6
1
6
P(A)=1-P(A’)=1- 5/8=3/8
P(AUB)=P(A) – P(B) – P(A B)
7/8=3/8+P(B) – ¼
P(B)=3/4
P(A B’)=P(A) – P(A B)
= 3/8-1/4
=1/8
A’ U B’ = (A B)’ and P(A’U B’) = 1 – P(A B) = 3/4
a).
b).
Only one of A, B occurs = (A B’)U((A’ B).
P(only one of A,Boccurs) = P(A B’)+P(A’ B)
= { P(A)-P(A B)} + { P(B)-P(A B)}
= 1/8 + ½ =5/8
E(X)=7/8
E(24X+6)=20
c).
E( 1-X)
a)
b).
Medium= 76.3 kg.
Interquartile range = 9 kg
7
50 35 th decile from curve .
Estimate of
10
60
50 30 th percentile from curve
Estimate of
100
e)
6).
1
1
c).
d).
1
1
2
=
0
(1 x)
1
2
1
3
( x )dx
2
5
109
REFERENCES
1) http://en.wikipedia.org/wiki/Statistics
2) A concise Course in A-Level Statistics By J. Crawshaw and J.Chambers, Stanley
Thornes Publishers, 1994
3) http://en.wikipedia.org/wiki/Probability
4) Business Calculation and Statistics Simplified, By N.A. Saleemi, 2000
5) http://microblog.routed.net/wp-content/uploads/2007/01/onlinebooks.html
6) Statistics: concepts and applications, By Harry Frank and Steven C Althoen,
Cambridge University Press, 2004
7) http://mathworld.wolfram.com/Statistics
8) http://mathworld.wolfram.com/Probability
9) probability Demystified, By Allan G. Bluman, McGraw Hill, 2005.
10) http://directory.fsf.org/math/
11) http://microblog.routed.net/wp-content/uploads/2007/01/onlinebooks.html
12) Lectures on Statistics, By Robert B. Ash, 2005.
13) Introduction to Probability, By Charles M. Grinstead and J. Laurie Snell, Swarthmore
College.
14) http://directory.fsf.org/math/
15) Simple Statistics, By Frances Clegg, Cambridge University Press 1982.
16) Statistics for Advanced Level Mathematics, By I. Gwyn Evans University College of
Wales, 1984.
18. Student Records
Module Developer Writing Tip. Though for most modules the final mark (for one module) will be closely linked to the summative
evaluation, it is often wise to mark or give points for the completion of other activities or formative evaluations. Module Developers
are therefore required to provide a clearly laid out “My Records” spreadsheet page that includes:
- Organized columns for entry of “future students” ;
- Organized columns for entry of all required marks ;
- Calculated columns to indicate overall achievement.
- Module Developers should provide the name of the EXCEL file.
Name of the EXCEL file :
Mathematics: Probability and Statistics Student Records
110
MAIN AUTHOR OF THE MODULE
Name: Mr. Paul Chege ( B.Ed(Sc), M.Ed )
Contact: paulamoud@yahoo.com
The module author is a teacher trainer at Amoud University,
Borama, Republic of Somaliland.
He has been a teacher trainer in Kenya, Republic of
Seychelles, and Somalia. He has been involved in
strengthening Mathematics and Sciences at secondary and
tertiary levels under the Japan International Corporation
Agency (JICA) programme in fifteen African countries.
He is married with three children.
20. File Structure
Module Developer Writing Tip. The file naming and structure must follow the AVU/PI Consortium template as defined and
explained by the AVU. Module Developers still need to provide the name of all the files (module and other files accompanying the
module).
Daily, each module will be loaded in the personal eportfolio created for each consultant. For this, training will be provided by
professor Thierry Karsenti and his team (Salomon Tchaméni Ngamo and Toby Harper).
Name of the module (WORD) file : Mathematics:
Probability and Statistics ( Word)
Name of all other files (WORD, PDF, PPT, etc.) for the module.
1. Mathematics: Probability and Statistics Student Records ( Excel)
2. Probability and Statistics: Marking Scheme for Summative Evaluation ( Word)
3. An Introduction to Probability and Random Processes, Textbook by Kenneth
Baclawski and Gian-Carlo Rota ( 1979) ( PDF)
4. Introduction To Probability, Textbook by Charles M. Grinstead and J. Laurie Snell
(PDF)
5. Lectures on Statistics, Textbook by Robert B. Ash (PDF)
111
